At the end you can simply apply the definition of logarithm... 2^x = 5 ---> x is the power to wich the base 2 must be raised in order to obtain 5 so you can write: x = log2(5)
I feel the quick way to approach this problem is to recognise that we have a polynomial where both y^3 and y exist. This should ring a bell that we should find some cubed number somewhere. We can write 130 = 125 + 5 = 5^3 + 5. That's it. Immediately we get y^3 - 5^3 + y - 5 = 0, or (y -5)(y^2 + 5y + 26) = 0, where y = 2^x. Here onwards, it is simple.
130=26×5=(5^2+1)×5=5^3+5; 8^x+2^x=(2^x)^3+(2^x); 2^x=5 My high school math teacher used to tell me,to understand an equation better you need to dis-simplify the brief side other than to simplify the complex side.
Me given IIT 20 yrs back and working in IT from 15+ yrs still willing to learn this so that I can teach my kid. Feels like back to square one. 😂 happy learning guys.
To all the pretentious people who keep commenting that she made it hard, please note that she is trying to teach everyone at any level, that's why she chose the most straightforward method that any type of student can follow. She is not teaching only the undiscovered geniuses such as yourselves.
@@christophersayrs907You’re such a weirdo, to take offence at the comment. He/she is right though, there are numerous ways to solve a problem in maths. Some ways will be faster than others, and may involve spotting a ‘shortcut’. But a teacher should always teach the ‘long’ method first, so that the student learns how to solve the problem in the standard way in the event that they cannot spot the shortcut.
FASTER APPROACH : 8^x+2^x is always increasing function, hence one root only. Setting x equal to some numbers, we realize that x is between 2 and 3. Consequently,y is between 4 and 8, but y is obviously less than 6 by analyzing cubic equation. So y is between 4 and 6, try 5 and we get that y=5. Everything can be solved in mind.
Change of bases can be used to simplify the final expression 8^log_2(5) = (2^log_2(5))^3 by associativity and commutativity =5^3 since x^log_x(a) = a =125 And the same with the other term, so 125+5 = 130
y3+y is stricly increasing, thus has only 1 real root, no need to complexify all what he did, like he did with all his base 2 and 5 logarithms. Just write 2^x=5 equals to x = ln(5)/ln(2), it's the basics of logs and exponentials
@@daakudaddy5453 test the monotonicity as the derivative of t³ = 3t² means that fnxn in increasing on R and as t approaches -infinity fnxn approaches -infinity and as fnxn approaches infinity fnxn approaches infinity which assured that there will be only one intersection point of x axis and fnxn hence only one possible solution log base 2 raised to 5 🙂 Duh!
We will also have complex values of x... Since we have a cubic polynomial y^3+y-130=0, it will therefore have 3 roots out of which 1 is real (log5/log2) and other two are complex Other two values of x hence will be: x=log(5+- isqrt71)/2/log2
I loved my Math's Teacher Sir A Hameed Wayn, in Metric Class ... And after him, you are the 2nd one, whom I would like to praise .. Since my First Teacher changed the School , and just for Mathematics I followed him to that school. From there, you can see my love for Mathematics . Liked your V-Log ... Though i dont know your name .
First observe that 8^x=(2^3)^x=(2^x)^3. Then set 2^x=y and you get the equation: y^3+y=y*(y^2+1)=130. A few trials gives y=5 and thus x=ln(5)/ln(2)=log(5)/log(2).
For all functions of type f(x)=ax^3+bx+c the equation f(x)=0 can have only one real root in case of a>0, b>0, because f'(x)=3ax^2+b>0 and therefore f(x) is monotonically increasing. Also, if the equation f(x)=0 has rational roots in the form p/q, p is a divisor of c and q is a divisor of a. If a=1, the rational roots are always whole numbers. One can immediately see that in the specific example f(y)=y^3+y-130=0 , y=5 is a solution, because we first check the whole divisors of 130 (+-1, +-2, +-5, +-10, +-13). Try to solve x^7+x^5+x^3+x=170 with your method....No chance. With above, you show that x=2 is the only root in just 2 rows.
It's too complicated, I mean after getting 2^x = 5 we get in accordance with the definition of a logarithm x = log2(5) where log2 is a logrithm with base 2. All following calculations in the video are unnecessary.
Having got to y³+ y = 130 it's not hard to try a few small numbers and see that y=5 is a solution. From that the quadratic part could be worked out, though it should be obvious that there can be no other real solutions since y>5 would be too big and y
I don't know anything about logarithms as it isn't in my syllabus but can we not solve it like this Note (8 to the power x is written as -8x) 8x +2x= 130 2(4x+1x) = 130 4x+1x=65 Anything as an exponent to 1 is 1 hence 4x+1=65 4x=64 4to the power x = 4 to the power 3 Hence x= 3
Before watching: This is not one that can be easily solved by simply plugging in integers and hoping for a result. Our solution is going to be somewhere between 2 (8^2 + 2^2 = 64+4=68) and 3 (8^3 + 2^3 = 512+8 = 520), and probably closer to 2 than to 3. Therefore, we have to actually do some calculations. Alright, 8 = 2^3, and (a^b)^c = a^(bc). Thus 8^x = (2^3)^x = 2^(3x). We declare U = 2^x. Then 8^x = 2^(3x) = u^3. Then we have u^3 + u = 130. Subtract 130 from both sides to get u^3+u-130=0. Now, we attempt to factor this. The factors of 130 are 1, 2, 5, 10, and 13. Of those, U=10 gives us results far too large, and U=2 gives us ones far too small. U=5 gives us 125+5-130 = 0, which is accurate. Thus, (u-5) is a factor. after doing some division, we can factor the equation into (u-5)(u^2 + 5u + 26 )=0. Going to use the quadratic formula on the second factor, we note that the discriminant is negative. Thus, these are not real roots, so we can skip this section. Thus, we will go with U=5. However, we're not done! That's the solution for U, not for X. U = 2^x. Thus, we have 2^x = 5. Take log_2 of both sides: Log_2(2^x) = log_2(5) -> X = log_2(5) Log_2 of 5 will be between log_2 of 4 (2) and log_2 of 8 (3), likely closer to 2 than 3. This checks out. If you want to change this to a different log using change of base, you may do so. log_b(a) = (log_c(a))/(log_c(b)),. Then using natural log ln, with a =5 and b = 2: x = (ln 5)/(ln2). (We're not doing this for the sake of precision, but rather so it can be easily checked on a calculator. A lot of calculators don't have functions built into them for logs of different bases, at least not ones that you can get to easily. Thus, you have to switch to either common log (log_10) or natural log (log_e))
Really crazy how easy this problem would be for my past self studying engineering. Being done with school and doing the same shit over and over again at work really rots your brain
@@Peter-Alexander yeah man I'm learning an additional 2 languages right now and also learning more music theory. Trading is also my side thing so I think the analytical part of my brain is still working to some extent. It's just that complex math isn't really my thing these days
I did it like this... 2^(3x)+2^x=130 Let 2^x=y Then y³+y=130 By observation, y=5 Hence 2^x=5 X=log(2)5, ie log 5 base 2 Overall, this problem was on the easier side if it's from olympiad, as I and my friends (Indian high-school students) were able to solve it pretty easily 😅
Hello. I am a mathematics enthusiast from Nigeria and I really want to increase my knowledge in mathematics. I would like to connect with you. Do you mind?
4:50 Redundant. By definition of logarithm, it is the value of the exponent to put on the base to get the argument of the logarithm. So basically in this case x is essentially, in base 2, log(5).
If you are asked for only the Real solution then youbare correct. If you want all solutions, you must take the quadratic into account and get two more Complex solutions.
When using common logs, complex solutions do not work. If you tried to substitute it back in the equation, the imaginary components would not cancel. Therefore, they are rejected.
Ótima explicação. Mas, quando você já havia encontrado que 2 elevado a x era igual a 5 , já poderia ter usado a definição de logaritmos e chegar direto na conclusão que x é igual a log de 5 na base 2 . Ou fazer assim seria um erro matemático? Parabéns pelo excelente vídeo!
In solution checking a^(log c to base a) can be written as c^(log a to base a) . So 8^(log 5 to base 2) can be written as 5^(log 8 to bas 2), which is 5^3 and 2^(log 5 to base 2), is 5^(log 2 to base 2) which is 5 5^3+5=130
To check the solution 2^x=5, or x=log5{base2}, For the original equation 8^x+2^x=130, rewrite as (2^x)3+2^x=130.substitute in the solution: 2^3(log5{base2})+2^log5{base2}=130. This becomes because of the laws of indices: 5^3+5=125+5=130.
130=26×5=(5^2+1)×5=5^3+5; 8^x+2^x=(2^x)^3+(2^x); 2^x=5 My high school math teacher used to tell me,to understand an equation better you need to dis-simplify the brief side other than to simplify the complex side.
great video but the verification can be a little bit better if it's like this: 8^log5 base 2 + 2^log5 base 2 = (2^3) ^log5 base 2 + 2^log5 base 2 = (2^log5 base 2) ^3 + 2^log5 base 2 as we already know a ^logN base a = N => (2^log5 base 2) ^3 + 2^log5 base 2 = 5 ^3 + 5 = 125 + 5 = 130 excellent video keep it up and upload more videos like this.
I think 5 is a rather easy to find "obvious" solution by searching a integer which cube is close to 130. Then it becomes a polynom division. The proposed factorisation is nevertheless very smart !
This is hard core algebra. Cubic equations, quadratic equations w/ the quadratic formula, logarithms, fractional exponents, ect…. I did stuff close to this level in high school. The difference was that it did not have multiple layers of this complexity. My takeaway from that experience was that algebra wasn’t hard so long as you worked a lot of different problems and got plenty of practice.
With the last digit of 130 being a zero, the last digit of 2^3x and 2^x must add to 10, 130 is an integer. The only pairs of numbers for x and 3x being powers of two would be 2,8, ,4,16 , 8,32 16,64 which sum to less than 130 or 32,128 etc,which which are greater than 130. your answer is approximate. 130 is discrete. This has no solution for exactly 130
reminder me my school years in one of the best math schools in Russia 25 years ago, now all forgotten but still these problems are solvable on the fly almost🙂 good times it was
Instead of that lengthy solution, let 2^x=y, and then let f(y)=y^3+y-130 For y=5, f(5)=0 f'(y)=3y^2+1, which is positive for all values of y, meaning f(y) is a monotonically increasing function, which makes y=5 the only root of f(y) Then 2^x=5 and solve using logarithmic properties
Yeah, but this is only shorter because it happened to be monotonically increasing. If it wasn’t, you would’ve done this step for no reason, and still had to do the lengthy solution.
Why aren't the imaginary solutions taken into account? I mean, they are solutions to the original equation, aren't they? Is there an assumption that we must find the real solutions only?
I think you could simply write 2^x = 5 => x = log2(5) without all these log divisions, because log2(5) literally means power to which we have to rase 2 in order to get 5, which is x in our case
I looked at this for like 2 minutes without a thought of any complex maths and thought the answer might be X= 2.25 and I’m honestly pretty pleased. Lol
I think the question is ill phrased because the types of the symbols in the equation are not specified. For example: no solution over integers. One solution over reals. 3 overvcomplex numbers. But what if we consider prime fields F_p with p>130 or any other algebra type where those symbols could be interpreted in?
y=5 is an obvious solution, taking all the terms on one side and differentiating, we get, f’(y) = 3y^2+1 > 0 for all y€R, hence f(y) has exactly one real root i.e. f(y=5) = 0.
This can be solved by vanishing factor method By putting y=5 y-5 is a factor 5^3+5-130=0 y^2(y-5)+5y(y-5)+26(y-5)=0 (y-5)(y^2+5y+26)=0 y-5=0 or y^2+5y+26=0 y=5. D for quadratic equation 5^2-4×1×26 is less than 0 No real roots. Solution will be y=5
As soon as she got it to Y Cubed plus Y = 130, the answer jumped out at me and I yelled it out loud, trying to beat her to the punch. Who knew I actuality had time to go check the mail, first?
This is an imperfect solution. I realized that the power I'd 8 could not be 3 abs 2 was too small, meaning it was not going g to be an integer. Kind of ridiculous in my book.
Nice, but you should see, that y^2+5y+26=0 can't have any real solutions without computing by seeing, that it's graph the shifted y^2-graph into positive direction.
Although I knew the answer to the problem, being lazy I preferred "Hit and Trial Method" from the given option for these type of questions. This saves a lot of time 😅😅
I could have just put some value in place and solved mathematically to make a guess that i have to go up or down a little. Than three attempts later i might have got x=2.3 I have done it many times. You dont need to go in multiple algebrac complications to solve a problem this simple. Afterall numbers are used to avoid hit and trial of actual objects. We can use numbers for hit and trial as that wont even use extra ink or time cz the equations are more complex than hit and trial. Smart work is also intelligence!
What happened in the end? Why the author used approximately solve? 2^log(2 5) = 5 by log's definition. And 8^log(2 5) = 2^3log(2 5) = (2^log(2 5))^3 = 5^3 = 125
Let y=2^x then y³+y-130=0 Easy to see y=5 is one solution. I will skip complex solutions which just requires synthetic division and quadratic formula to solve. 5=2^x -> log_2(5)= x -> x= ln5/ln2
Lol, I've done that independently from UA-cam thumbnail and my result was X = 1 / [ log130(10) ] which is X = log10(130) which reads as [ log 130 to base 10 ] Edit: It was quite close as my result evaluates to 2.1139 and the actual result evaluates to 2.3219
y³+y-130 = 0 y³-125+y-5=0 (y-5)(y²+5y+25)+(y-5)=0 (y-5)(y²+5y+26)=0 The real solution is y=5 and from the general 2nd grade formula there will be 2 complex solutions.
I do not yet have the mathematical experience to have come up with that line of thinking to jump to thinking of the factorisation of 130 then rewriting as factor by grouping. I got to the step where I substituted u = 2^x but had no idea that was actually the way to proceed. I was stuck at what to do now with the 130 as I had no obvious way to factorise u^3 + u - 130 = 0.
Computing 2.32 to the equation does not equals both the sides. There is a significant difference of around .51 in both sides. 8^2.32 = 124.49 + 2^2.32 = 4.99 = 129.48 ≠ 130. Even in points these should be equal. Consider this 4^x = 8x. If the value of x = .154953 and computing it in equation both the sides are equal. 1.2396 = 1.2396. In your solution it is not the case. Rounding off does not goes well with the equation which has to make both sides equal. Computing this interdisciplinary tolerance ≠ tulerance. Spellings matters, even one alphabet can make the entire spelling wrong. Tolerance = tolerance ≠ tulerance just as 130 = 130 ≠ 129.48 Now, try this. 8^ √ x +2^x = 80. Find the value of x. Much simpler but demonstrates the equality perfectly. Your solution is approximately equal(≈) not equal(=) to 130. Hence, it demonstrates ≈ and not =.
I got from a much easier and faster way 2*3x and 2*x should be equal to 130 so the sum of two of 2 power something should be equal to 130 which is close to 2*7 hence if x=2 the answer is 64+4 and if x= 3 it would be 528, therefore, x should be between 2 and 3 and closer to 2 than 3 and 3x kinda feels be close to 7 as 128 is close to 130 so 7/3=2.33 and as it wouldn't add up (cuz 127+4,9 is greater than 130 we should decrease 2,33 by 0,01 per time to find the right answer then we got the answer of 2.32
You make the solution to obtain the real part seem too complicated. from the point where (2^x)=5 by just using the conversion law from Exponents to log we obtain Log5_2 = x. Even to check the result, using logs instead the decimal produces a more exact result or (125+5=130).
At the end you can simply apply the definition of logarithm...
2^x = 5 ---> x is the power to wich the base 2 must be raised in order to obtain 5
so you can write:
x = log2(5)
No, she has to do a 5-minute thesis on it (to go from 2^x=5 to x=log5 to base 2, otherwise her students might not understand it.
@@tintiniitkwe learned it as logba=e where e is the exponent b is the base and a the “answer”. its very helpful
😂😂
I think you are right
I feel the quick way to approach this problem is to recognise that we have a polynomial where both y^3 and y exist. This should ring a bell that we should find some cubed number somewhere. We can write 130 = 125 + 5 = 5^3 + 5. That's it. Immediately we get
y^3 - 5^3 + y - 5 = 0, or (y -5)(y^2 + 5y + 26) = 0, where y = 2^x.
Here onwards, it is simple.
(8×15)+(2×5)=130
thats what i did, i got hung up because i did log(5)-log(2) instead of log2(5)
Wow. This was way easier! I should've just thought through some numbers and would've hit it just by thinking of those 1st few perfect cubes.
y=2x , find the value idiot
She made it way harder to solve than she had to.
130=26×5=(5^2+1)×5=5^3+5;
8^x+2^x=(2^x)^3+(2^x);
2^x=5
My high school math teacher used to tell me,to understand an equation better you need to dis-simplify the brief side other than to simplify the complex side.
Wowwww😮😮😮
Nice! I did it like this as well or, similar.
Yeah, it just depends. A lot of times that advice will work, but not always.
Yours can't guarantee there is unique root.
@@zhenyuzhai4098 I know it should be unique in this case
This is why I'm going back to watching cooking videos.
Or funny cats
Come on. Everyone should know how to solve this
@@Aman-nk5uq he means it's too easy
Most mathematicians probably can't cook worth a damn. Most people can't.
😂😂 I don't blame you
All these problems seem a little too easy for olympiads
Are you sure?
@@estefanocrespo7930yes these are like middle school problems for us
@@estefanocrespo7930im not him but yeah this is pretty easy for olympiads because i can even answer it
Maybe it's the special olypiad
😂😂😂😂
Instead of using decimals you could have used more logarithmic identities when you were checking. I think that would have been cleaner.
Came here to note that.
Exactly.. 2^(log 125/log 2) = 125 and so on.
@@manojpadmanabhan2615 let's be nice with our math mates
Me given IIT 20 yrs back and working in IT from 15+ yrs still willing to learn this so that I can teach my kid. Feels like back to square one. 😂 happy learning guys.
Looks like you're stuck in a loop in life.
Don't teach to kids whatever that's not useful. Times are changing rapidly. Pretty sure you never used this in your working life.
*for 15 years
@@krishmavthese type of outside the box question are what asked in sof olympiads
@@ankurmondal3220so you are studying fkr sof olympiads, go and give the real IMO
Make sure you don't piss in your pants
To all the pretentious people who keep commenting that she made it hard, please note that she is trying to teach everyone at any level, that's why she chose the most straightforward method that any type of student can follow. She is not teaching only the undiscovered geniuses such as yourselves.
@@christophersayrs907You’re such a weirdo, to take offence at the comment. He/she is right though, there are numerous ways to solve a problem in maths. Some ways will be faster than others, and may involve spotting a ‘shortcut’. But a teacher should always teach the ‘long’ method first, so that the student learns how to solve the problem in the standard way in the event that they cannot spot the shortcut.
FASTER APPROACH : 8^x+2^x is always increasing function, hence one root only. Setting x equal to some numbers, we realize that x is between 2 and 3. Consequently,y is between 4 and 8, but y is obviously less than 6 by analyzing cubic equation. So y is between 4 and 6, try 5 and we get that y=5. Everything can be solved in mind.
Change of bases can be used to simplify the final expression
8^log_2(5) = (2^log_2(5))^3 by associativity and commutativity
=5^3 since x^log_x(a) = a
=125
And the same with the other term, so
125+5 = 130
I agree, the finish could have been done a bit more elegantly with out finding the decimal approx.
Exactly, I was like what is she doing....this can't be an Olympiad question
Incredible how much effort you put in this. 130=5³+5¹ and this is identical to your y³+y. Therefore y = 2^x = 5. => x = log2 (5). Easy.
It’s just to prove that there is no other solutions with real values
You also have to check for other possible solutions. Duh!
y3+y is stricly increasing, thus has only 1 real root, no need to complexify all what he did, like he did with all his base 2 and 5 logarithms.
Just write 2^x=5 equals to x = ln(5)/ln(2), it's the basics of logs and exponentials
@@daakudaddy5453 test the monotonicity as the derivative of t³ = 3t² means that fnxn in increasing on R and as t approaches -infinity fnxn approaches -infinity and as fnxn approaches infinity fnxn approaches infinity which assured that there will be only one intersection point of x axis and fnxn hence only one possible solution log base 2 raised to 5 🙂 Duh!
We will also have complex values of x...
Since we have a cubic polynomial y^3+y-130=0, it will therefore have 3 roots out of which 1 is real (log5/log2) and other two are complex
Other two values of x hence will be: x=log(5+- isqrt71)/2/log2
The imaginary solutions were extraneous. That's why they were explicitly discarded.
@@NicholasOfAutrecourtextraneous or extra anus?
I loved my Math's Teacher Sir A Hameed Wayn, in Metric Class ... And after him, you are the 2nd one, whom I would like to praise ..
Since my First Teacher changed the School , and just for Mathematics I followed him to that school. From there, you can see my love for Mathematics . Liked your V-Log ... Though i dont know your name .
I remember fondly a time where maybe, just maybe, I might have had some idea, but I haven’t used anything beyond grade 9 algebra in 20 years.
First observe that 8^x=(2^3)^x=(2^x)^3. Then set 2^x=y and you get the equation: y^3+y=y*(y^2+1)=130. A few trials gives y=5 and thus x=ln(5)/ln(2)=log(5)/log(2).
Just subtract 130 from both sides and solve for the x-intercepts of the equation 8^x + 2^x - 130 = y. When y= 0 the graph intercepts the x axis.
For all functions of type f(x)=ax^3+bx+c the equation f(x)=0 can have only one real root in case of a>0, b>0, because f'(x)=3ax^2+b>0 and therefore f(x) is monotonically increasing. Also, if the equation f(x)=0 has rational roots in the form p/q, p is a divisor of c and q is a divisor of a. If a=1, the rational roots are always whole numbers. One can immediately see that in the specific example f(y)=y^3+y-130=0 , y=5 is a solution, because we first check the whole divisors of 130 (+-1, +-2, +-5, +-10, +-13).
Try to solve x^7+x^5+x^3+x=170 with your method....No chance. With above, you show that x=2 is the only root in just 2 rows.
It's too complicated, I mean after getting 2^x = 5 we get in accordance with the definition of a logarithm x = log2(5) where log2 is a logrithm with base 2. All following calculations
in the video are unnecessary.
Yeah, exactly my thoughts: just use the definition!
It was painful to see how she derives x after getting 2^x = 5 (((
Having got to y³+ y = 130 it's not hard to try a few small numbers and see that y=5 is a solution. From that the quadratic part could be worked out, though it should be obvious that there can be no other real solutions since y>5 would be too big and y
Yes I think so
Два графика функций
у=х^3 и у=-х+130 пересекаются в одной точке в 1 четверти=> х=5 единственное решение
Confuse the number 26y - 25y, can you explaine it!
26y-25y=y
A doctor here , thanking my stars for me choosing biology and not maths😅😅
I don't know anything about logarithms as it isn't in my syllabus but can we not solve it like this
Note (8 to the power x is written as -8x)
8x +2x= 130
2(4x+1x) = 130
4x+1x=65
Anything as an exponent to 1 is 1 hence
4x+1=65
4x=64
4to the power x = 4 to the power 3
Hence x= 3
Before watching:
This is not one that can be easily solved by simply plugging in integers and hoping for a result. Our solution is going to be somewhere between 2 (8^2 + 2^2 = 64+4=68) and 3 (8^3 + 2^3 = 512+8 = 520), and probably closer to 2 than to 3.
Therefore, we have to actually do some calculations.
Alright, 8 = 2^3, and (a^b)^c = a^(bc). Thus 8^x = (2^3)^x = 2^(3x).
We declare U = 2^x. Then 8^x = 2^(3x) = u^3.
Then we have u^3 + u = 130.
Subtract 130 from both sides to get u^3+u-130=0.
Now, we attempt to factor this. The factors of 130 are 1, 2, 5, 10, and 13.
Of those, U=10 gives us results far too large, and U=2 gives us ones far too small. U=5 gives us 125+5-130 = 0, which is accurate.
Thus, (u-5) is a factor. after doing some division, we can factor the equation into (u-5)(u^2 + 5u + 26 )=0.
Going to use the quadratic formula on the second factor, we note that the discriminant is negative. Thus, these are not real roots, so we can skip this section.
Thus, we will go with U=5. However, we're not done! That's the solution for U, not for X.
U = 2^x. Thus, we have 2^x = 5.
Take log_2 of both sides:
Log_2(2^x) = log_2(5) -> X = log_2(5)
Log_2 of 5 will be between log_2 of 4 (2) and log_2 of 8 (3), likely closer to 2 than 3. This checks out.
If you want to change this to a different log using change of base, you may do so. log_b(a) = (log_c(a))/(log_c(b)),.
Then using natural log ln, with a =5 and b = 2:
x = (ln 5)/(ln2).
(We're not doing this for the sake of precision, but rather so it can be easily checked on a calculator. A lot of calculators don't have functions built into them for logs of different bases, at least not ones that you can get to easily. Thus, you have to switch to either common log (log_10) or natural log (log_e))
Superb
☠️
For 2^x=y
y³+y=130
y(y²+1)=130
Clearly for y=5, equation satisfied so x=log5(base 2) is the correct answer.
How does this relate to real world practicality, thank you.
Really crazy how easy this problem would be for my past self studying engineering. Being done with school and doing the same shit over and over again at work really rots your brain
Это не легкая задача
Challenge your brain in your free time or find a more interesting job (when possible) 😊
@@Peter-Alexander yeah man I'm learning an additional 2 languages right now and also learning more music theory. Trading is also my side thing so I think the analytical part of my brain is still working to some extent. It's just that complex math isn't really my thing these days
I did it like this...
2^(3x)+2^x=130
Let 2^x=y
Then y³+y=130
By observation, y=5
Hence
2^x=5
X=log(2)5, ie log 5 base 2
Overall, this problem was on the easier side if it's from olympiad, as I and my friends (Indian high-school students) were able to solve it pretty easily 😅
Hello.
I am a mathematics enthusiast from Nigeria and I really want to increase my knowledge in mathematics. I would like to connect with you.
Do you mind?
Same thoughts here. I literally solved it in mind using the similar way as you in 1min
Yeah but try to solve it when u were 13 or 15.
Damn we really dont need a 8 minute to solve this one. Nice observation.
people @13-15 in india can solve this way faster compared to the older folks in india btw@@m3zuss
4:50
Redundant. By definition of logarithm, it is the value of the exponent to put on the base to get the argument of the logarithm. So basically in this case x is essentially, in base 2, log(5).
If you are asked for only the Real solution then youbare correct. If you want all solutions, you must take the quadratic into account and get two more Complex solutions.
Don't be clever. When is a maths solution going to ask for complex solutions
@@planomathandscienceanywhere above high school, and at high school levels in some countries, expect a full answer unless otherwise specified.
yeah! I want all answers!
When using common logs, complex solutions do not work. If you tried to substitute it back in the equation, the imaginary components would not cancel. Therefore, they are rejected.
@1234larry1 So apparently, Wolfram Alpha is wrong as well? It can substitute the values back in without issue.
y(y²+1) = 5(5²+1)
...at this stage it should be apparent that y = 5 🤪
Given y = 2^x
Then 2^x = 5
log²(2^x) = log²(5)
x = log²(5)
Ótima explicação. Mas, quando você já havia encontrado que 2 elevado a x era igual a 5 , já poderia ter usado a definição de logaritmos e chegar direto na conclusão que x é igual a log de 5 na base 2 .
Ou fazer assim seria um erro matemático?
Parabéns pelo excelente vídeo!
I also wondered why we need last manipulations with Log, it is redundant. 2^X=Y. X = logY.
In solution checking
a^(log c to base a) can be written as c^(log a to base a) .
So 8^(log 5 to base 2) can be written as 5^(log 8 to bas 2), which is 5^3 and 2^(log 5 to base 2), is 5^(log 2 to base 2) which is 5
5^3+5=130
To check the solution 2^x=5, or x=log5{base2}, For the original equation 8^x+2^x=130, rewrite as (2^x)3+2^x=130.substitute in the solution: 2^3(log5{base2})+2^log5{base2}=130. This becomes because of the laws of indices: 5^3+5=125+5=130.
This is why I know anatomy so well!
Let a=2^x, then a^3+a=130, so a=5 is a solution, then x= ln5/ln2. The other solutions can be found by factoring out (a-5) and solving the quadratic.
Factors of -130 according to you are 26 and -5. Why did you use 26 and-25.?
Because y = 26y - 25y. If she used 5 instead of 25, it will be 26y - 5y = 21y
125+5=130
125=5^3.
Just plugged in first "obvious" solution to reduce cubic to quadratic equation.
130=26×5=(5^2+1)×5=5^3+5;
8^x+2^x=(2^x)^3+(2^x);
2^x=5
My high school math teacher used to tell me,to understand an equation better you need to dis-simplify the brief side other than to simplify the complex side.
-130 has not been factorised..only “y” at second place has been extended “26y-25y”..
She did not explain it. It’s better to just notice 5 is a root and then divide the equation by x-5
Спасибо! Не понимаю концовку с таймкода 4:38, и так ясно, что логорифм - это степень числа по основанию.🧐
1:37 is confusing? Why and how can you write y=26y-5y?
У = 26у - 25у
У = 1у
@@КристиянСлавчев-ф7рhow does 26&25 relate to 130
great video but the verification can be a little bit better if it's like this:
8^log5 base 2 + 2^log5 base 2 = (2^3) ^log5 base 2 + 2^log5 base 2
= (2^log5 base 2) ^3 + 2^log5 base 2
as we already know a ^logN base a = N
=> (2^log5 base 2) ^3 + 2^log5 base 2 = 5 ^3 + 5
= 125 + 5
= 130
excellent video keep it up and upload more videos like this.
In Turkey you learn solving these at the age of 14-15 already. This is easy peasy and has nothing to do with olympics.
Stop at 1:15. You kind of know that Y has to be an integral number given it's a math Olympiad problem. A few guess will gets Y=5 easily.
Use rational zero theorem.
Y=5.
And divide the function by (y-5) and get the quadratic.
Much faster....
Inspecting, 2
2^log²5 = 5 (by definition)
8^log²5 = (2^3)^log²5=
(2^log²5)^3=
5^3=125
There is no need to do approximete calculations.
In Italy this is called "col senno di poi"! 😂
The longest way to do that equation.
I think 5 is a rather easy to find "obvious" solution by searching a integer which cube is close to 130. Then it becomes a polynom division.
The proposed factorisation is nevertheless very smart !
This is hard core algebra. Cubic equations, quadratic equations w/ the quadratic formula, logarithms, fractional exponents, ect…. I did stuff close to this level in high school. The difference was that it did not have multiple layers of this complexity. My takeaway from that experience was that algebra wasn’t hard so long as you worked a lot of different problems and got plenty of practice.
I'm already confused at 1:29 . Where did the 25 come from?
130 = 26х5
In real exams, you can just substitute the given choices if satisfies the 130. It will save lot of time instead of solving.
In real exams you are not given choices. What educational system did you attend? "Fast food and exams Inc."?
@@AceGunner72 maybe you didnt take any licensure exams.
Use substitution. Substitute 2^x =y and solve polynomial equation.
With the last digit of 130 being a zero, the last digit of 2^3x and 2^x must add to 10, 130 is an integer. The only pairs of numbers for x and 3x being powers of two would be 2,8, ,4,16 , 8,32 16,64 which sum to less than 130 or 32,128 etc,which which are greater than 130. your answer is approximate. 130 is discrete. This has no solution for exactly 130
reminder me my school years in one of the best math schools in Russia 25 years ago, now all forgotten but still these problems are solvable on the fly almost🙂 good times it was
Какая школа?
Instead of that lengthy solution, let 2^x=y, and then let f(y)=y^3+y-130
For y=5, f(5)=0
f'(y)=3y^2+1, which is positive for all values of y, meaning f(y) is a monotonically increasing function, which makes y=5 the only root of f(y)
Then 2^x=5 and solve using logarithmic properties
Yeah, but this is only shorter because it happened to be monotonically increasing. If it wasn’t, you would’ve done this step for no reason, and still had to do the lengthy solution.
I have simplier solution. Multiply both by root x, so then 8+2 = x base root of 130. Which easily = 2.3....
I think the easiest solution is that we know 2^7=128 next is 8^(1/3) = third root 8 =2 so 128+2=130
This was my solution as well. Much simpler.
Both the exponents are “x” so the value needs to be same. Can’t be 7 for one x and 1/3 for the other
Why aren't the imaginary solutions taken into account? I mean, they are solutions to the original equation, aren't they? Is there an assumption that we must find the real solutions only?
I think you could simply write 2^x = 5 => x = log2(5) without all these log divisions, because log2(5) literally means power to which we have to rase 2 in order to get 5, which is x in our case
Was in it better to approximate x between 2 and 3 at a glance?
I looked at this for like 2 minutes without a thought of any complex maths and thought the answer might be X= 2.25 and I’m honestly pretty pleased. Lol
I also have zero training in mathematics outside of high school 10+ years ago so go very easy on me
I think the question is ill phrased because the types of the symbols in the equation are not specified. For example: no solution over integers. One solution over reals. 3 overvcomplex numbers. But what if we consider prime fields F_p with p>130 or any other algebra type where those symbols could be interpreted in?
y=5 is an obvious solution, taking all the terms on one side and differentiating, we get, f’(y) = 3y^2+1 > 0 for all y€R, hence f(y) has exactly one real root i.e. f(y=5) = 0.
8^x+2^x=(2^3)^x+2^x=(2^x)^3+2^x, let 2^x=a, a^3+1=130, a^3=129, a=4.971, 2^x=4.971, x=ln4.971/ln2,x= 2.32
This can be solved by vanishing factor method
By putting y=5
y-5 is a factor
5^3+5-130=0
y^2(y-5)+5y(y-5)+26(y-5)=0
(y-5)(y^2+5y+26)=0
y-5=0 or y^2+5y+26=0
y=5.
D for quadratic equation
5^2-4×1×26 is less than 0
No real roots.
Solution will be y=5
As soon as she got it to Y Cubed plus Y = 130, the answer jumped out at me and I yelled it out loud, trying to beat her to the punch. Who knew I actuality had time to go check the mail, first?
Yeah I got it too. But when you check it, you should not convert to log value and you can simply find it.
Thanks you so much for refreshing my memory from 10 years ago, Now I wish I have continued on the math field instead
Esa es una forma extremadamente larga.resolverlo, basta factorizar al principio por 2 elevado a x y después aplicar logaritmo, se resolvía en 2 pasos
Yes 130=26x5, but how to get y=26y-25y ?
I mean how to think of such an inference ?
Experience :)
26*5=130
26*25=650
I think solution is Not correct
This is an imperfect solution. I realized that the power I'd 8 could not be 3 abs 2 was too small, meaning it was not going g to be an integer. Kind of ridiculous in my book.
So hard. Congrats. Several math concepts 👏👏👏👏👏
Log 2^x = x log 2. Please learn this. For your solution:
X= log5/log2.
Nice, but you should see, that y^2+5y+26=0 can't have any real solutions without computing by seeing, that it's graph the shifted y^2-graph into positive direction.
very informative and easy way of teaching
Go easier by using Factorizing with polynomial and basic logarithm
Although I knew the answer to the problem, being lazy I preferred "Hit and Trial Method" from the given option for these type of questions. This saves a lot of time 😅😅
Nicely done. Weird how your 2's are written differently even in the same equation
I could have just put some value in place and solved mathematically to make a guess that i have to go up or down a little. Than three attempts later i might have got x=2.3
I have done it many times. You dont need to go in multiple algebrac complications to solve a problem this simple. Afterall numbers are used to avoid hit and trial of actual objects. We can use numbers for hit and trial as that wont even use extra ink or time cz the equations are more complex than hit and trial. Smart work is also intelligence!
Why aren't we considering the non-real solutions? They're pretty straightforward too!
1st: x=log2(5)
2nd: x=log2(-5/2 - 1/2 (i sqrt(79)))
3rd: x=log2(-5/2 + 1/2 (i sqrt(79)))
Pretty easy!!!
What happened in the end? Why the author used approximately solve? 2^log(2 5) = 5 by log's definition. And 8^log(2 5) = 2^3log(2 5) = (2^log(2 5))^3 = 5^3 = 125
Your voice is so soothing
Let y=2^x then y³+y-130=0
Easy to see y=5 is one solution. I will skip complex solutions which just requires synthetic division and quadratic formula to solve.
5=2^x -> log_2(5)= x -> x= ln5/ln2
Lol, I've done that independently from UA-cam thumbnail and my result was X = 1 / [ log130(10) ] which is X = log10(130) which reads as [ log 130 to base 10 ]
Edit: It was quite close as my result evaluates to 2.1139 and the actual result evaluates to 2.3219
y³+y-130 = 0
y³-125+y-5=0
(y-5)(y²+5y+25)+(y-5)=0
(y-5)(y²+5y+26)=0
The real solution is y=5 and from the general 2nd grade formula there will be 2 complex solutions.
Aah my favourite maths algebra those days ❤🎉
I do not yet have the mathematical experience to have come up with that line of thinking to jump to thinking of the factorisation of 130 then rewriting as factor by grouping. I got to the step where I substituted u = 2^x but had no idea that was actually the way to proceed. I was stuck at what to do now with the 130 as I had no obvious way to factorise u^3 + u - 130 = 0.
To complicate solution
We can divide by 2^x and y= 2^×
Your handwriting and logical thinking ability are awesome ❤
Computing 2.32 to the equation does not equals both the sides. There is a significant difference of around .51 in both sides. 8^2.32 = 124.49 + 2^2.32 = 4.99 = 129.48 ≠ 130. Even in points these should be equal. Consider this 4^x = 8x. If the value of x = .154953 and computing it in equation both the sides are equal. 1.2396 = 1.2396. In your solution it is not the case. Rounding off does not goes well with the equation which has to make both sides equal.
Computing this interdisciplinary tolerance ≠ tulerance. Spellings matters, even one alphabet can make the entire spelling wrong. Tolerance = tolerance ≠ tulerance just as 130 = 130 ≠ 129.48
Now, try this. 8^ √ x +2^x = 80. Find the value of x. Much simpler but demonstrates the equality perfectly.
Your solution is approximately equal(≈) not equal(=) to 130. Hence, it demonstrates ≈ and not =.
I got from a much easier and faster way 2*3x and 2*x should be equal to 130 so the sum of two of 2 power something should be equal to 130 which is close to 2*7 hence if x=2 the answer is 64+4 and if x= 3 it would be 528, therefore, x should be between 2 and 3 and closer to 2 than 3 and 3x kinda feels be close to 7 as 128 is close to 130 so 7/3=2.33 and as it wouldn't add up (cuz 127+4,9 is greater than 130 we should decrease 2,33 by 0,01 per time to find the right answer then we got the answer of 2.32
4:47 x is already log 5 to the base 2 as per logarithm definition,. Isnt it
..am I the only one to solve it in their mind after seeing the thumbnail?
yes. You're the only one.
Its a mid level math problem for college exam in Turkey.
a=2^x
a^3 + a = 130
both increasing
a=5 is the only solution
2^x = 5
x = log 2 (5)
:/
İ know short answer for this question and it is very easy ,
Beautiful question. It's answer is log5/log2 both on the base 10 👌👌
or any other base for that matter
SInce this is a cubic it must have a trivial solution since no math olympiad participant would be able to solve a proper cubic analytically
@4:10 - no real roots? That's why we have complex numbers! Rework the video and give us two more correct answers!
You make the solution to obtain the real part seem too complicated. from the point where (2^x)=5 by just using the conversion law from Exponents to log we obtain Log5_2 = x. Even to check the result, using logs instead the decimal produces a more exact result or (125+5=130).
redundant explanation from 4:49.
From logarithm definition we know that it is log_2(5)