Can You Pass Cambridge Entrance Exam?
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- Опубліковано 28 вер 2024
- Entrance examination and Math Olympiad Question in 2020. If you're reading this ❤️.
What do you think about this problem?
Hello My Friend ! Welcome to my channel. I really appreciate it!
@higher_mathematics
#maths #math
![The most beautiful equation in math, explained visually [Euler’s Formula]](/img/n.gif)
With Cardano/Tartaglia formula we get the following reasonning:
8^x + 2^x = 130
(2^3)^x + 2^x = 130
(2^x)^3 + 2^x = 130
let k = 2^x
k^3 + k = 130
k^3 + k - 130 = 0
k^3 + k - 130 = 0 is based on k³ + kp + q = 0 template, with:
• p = 1
• q = -130
recall and application of the Cardano/Tartaglia formula:
• k = [-q/2 + √(q²/4 + p³/27)]^(1/3) + [-q/2 - √(q²/4 + p³/27)]^(1/3)
k = (-(-130)/2 + √(((-130)^2)/4 + ((1)^3)/27))^(1/3) +
(-(-130)/2 - √(((-130)^2)/4 + ((1)^3)/27))^(1/3)
k = 5
2^x = k = 5
2^x = 5
x = ln(5)/ln(2)
x = 2.321928
----- check -----
8^x + 2^x = 130
8^2.321928 + 2^2.321928 = 129.999997
----- final results -----
■ x = ln(5)/ln(2)
■ x = 2.321928
🙂
too long , when you let k = 2^x,
let f(k) k^3+k-130=0
just find the factor of f(k),
that's 5 ,(when k=5,f(k)=0),so that k-5 is the factor of f(k),
use the long division method,
we get
(k-5)(k^2+5k+26)=0
k=5,k= (-5+/- sqrt(79)i)/2
2^x =5
x=log(2.,5)
x=2.32 (3 con fig)
@@anson2075same thing really, cardano is more general to apply to other problem.
Though a faster is observe x=5 is a solution of x^3 + x = 130, because x^3 + x is strictly monotonic so x=5 is the only real solution
So no complex values??! Huh?
@@johnsmith1953x Hi Sir, Below is the continuation as per your expectation...
/// euclidean division
• from k^3 + k - 130 = 0 we got k = 5
• k^3 + 0·k^2 + k - 130 = 0
• (k^3 + 0·k^2 + k - 130) / (k - 5) = (k^2 + 5k + 26)
/// simple quadratic equation
• k^2 + 5k + 26 = 0
• Δ = 5² - 4·1·26 = 25 - 104 = -79
• √Δ = ±i√79
-> root #1: k = (-5 + i√79)/(2·1) = -5/2 + i√79/2
-> root #2: k = (-5 - i√79)/(2·1) = -5/2 - i√79/2
/// root #1: k = -5/2 + i√79/2
1) mod of k:
• mod = √[(5/2)^2 + (√79/2)^2]
• mod = √(25/4 + 79/4)
• mod = √(104/4)
• mod = √26
2) arg of k:
• arg(k) = π - arctan(√79/5)
3) exponential form of k:
• k = √26·e^i·(π - arctan(√79/5))
4) recall: k = 2^x:
• 2^x = k = √26·e^i·(π - arctan(√79/5))
• 2^x = √26·e^i·(π - arctan(√79/5))
• x = ln(√26·e^i·(π - arctan(√79/5)))/ln(2)
/// root #2: k = -5/2 - i√79/2
1) mod of k:
• mod = √[(5/2)^2 + (√79/2)^2]
• mod = √(25/4 + 79/4)
• mod = √(104/4)
• mod = √26
2) arg of k:
• arg(k) = π - arctan(√79/5)
3) exponential form of k:
• k = √26·e^i·(arctan(√79/5) - π)
4) recall: k = 2^x:
• 2^x = k = √26·e^i·(arctan(√79/5) - π)
• 2^x = √26·e^i·(arctan(√79/5) - π)
• x = ln(√26·e^i·(arctan(√79/5) - π))/ln(2)
/// final results
■ x = ln(5)/ln(2)
■ x = ln(√26·e^i·(π - arctan(√79/5)))/ln(2)
■ x = ln(√26·e^i·(arctan(√79/5) - π))/ln(2)
🙂
Oh yeah sure because everyone remembers the cubic formula just like that. 😂
Once we had y^3 + y - 130 = 0, it was immediately clear that y = 5 and the problem was solved. I’m not sure that the approach of changing y to 26y - 25y is an easier approach; it seems it requires more insight. Your thoughts please.
There are two more solutions, though complex. I agree that if you can see the (y-5) factor, it’s probably faster to just do the poly division.
You divide the polynomial with y-5 and check the remainder, which is y²+5y+126. It's clearly a polynomial division issue. You should check the roots of the remainder because they could be real numbers too. So in general a solution other than 5 could be also feasible.
Ergo: x = ln(5)/ln(2)
@@jjeanniton why using a base other than 2? It is clearly a base 2 logarithm: log2(5)
Yep
From 2^x = 5, x = log 5 to the base 2 directly follows. No need to do any algebraic manipulation.
Yes I also did same
He uses the theorem to prove the definition of logarithm.
For some reason they dothis shit all the time in school
@@beaumatthews6411 This question was undoubtedly first written long before calculators were invented, hence the simple solution x=log(5)/log(2)
The answer would never be expected in log base 2 because the log tables used before calculators only had log base 10 (log) and natural log (ln). The final answer would be obtained using a slide rule. Fun methods that put a man on the Moon.
I was going to say that. Cheers!
A guess is a guess. Without demonstrating thinking process, replacing y = 26y - 25y is nothing but a guess, which teaches no general approach to solving similar problems. It is no better that noticing that y^3 + y grows pretty fast, so we only need to try a few integers to get 130. And 5 pops up almost instantly. It’s also a guess, but much more natural and less contrived.
agree with you, applying y = 25y - 25 y is far from a demonstration process, it is simply a trick. I prefer to use my brain and notice, as you did, that y^3+y grows fast
Isn't that what mid term split is all about tho? We deliberately choose those factors that gives the constant on multiplying and the coefficient of the lesser degree variable in the equation on adding?
equations can have irrational, complex and non integer roots, which are often impossible to guess (although they can be approximated). What you made was a guess for an integer root, and its standard practice to check a cubic for an integer root first, and then use factor thereom. Sometimes, an integer root does not exist or is too large and you must use cleverer methods. You often need to find the complex roots (which were skipped over by the poster)
It is not contrived to guess a particular factorisation. A lot of good algebra problems are about making clever observations, factorizations, splitting etc. Many good problems asked in olympiade and competetive exams all around the world are based on using clever little observations, which instantly simplify the problem.
There are 'non contrived' ways to solve cubics, but they are absurdly long (check out the cubic formula. The derivation is pretty interesting). And they do not even exist for a polynomial equation greater than 4 degree.
Being a middle school student i thought that this is an algebra equation in one variable😂😂. I thought it is 8x + 2x = 130 and here x = 13 i was like it is so damn easy why are they freaking out😂😂😂
Interesting video, but if you're going to write the product as 26*5, why not just write it as 25*5 + 5 since you have y^3 + y, so 125+5 = y^3 + y, 5^3 + 5 = y^3 + y therefore y = 5 x = log_2(5).
Easy ques and i m going to Cambridge...wish me luck❤
I have graduated in Mechanical Engineering from RUT in Poland. 32 years ago. And I solved this problem in three lines. :) Applying logarithms.
Good Luck! :D
@@piotrstrzelczyk5013 Thanks mate :)
@@piotrstrzelczyk5013 same but i keep getting 2.
I changed 130 to 2⁷ + 2. Changed the other side to powers of 2 as well
Took log, then cancelled
And ended up with 4x = 8
You could substitute y=2**x then use the rational root theorem to find the factor (x-5)
Much better.
thats what I did lol
By observation from y^3+y-130=0 we get y=5 so x=(ln5/ln2) which verifies the initial equation
Eureka, my insomnia has been cured. (Fell asleep at 4 min.)
I tighten bolts and signal cranes for a living and got this inside of about 3 seconds. Cambridge should really tighten up their entrance requirements.
Well your overqualified for your job then and should look for a job that requires more intellect than tightening bolts because only 1% of adults in the general population could solve this problem.
You tighten bolts and signal cranes for a living, but that doesn't mean that you aren't a math genius too!
@@charlesstimler9276 Arithmetic, geometry, and algebra maybe. I have zero clue about calculus
Interesting.
The solution hinges on writing 130 = 5 x 26, which is noted and exploited in the on-screen "working out".
Similarly observed: 26 = 25 + 1 = 5^2 + 1.
Hence 130 = 5 x (5^2 + 1) = 5^3 + 5.
Comparing 5^3 + 5 = 130 with 8^x + 2^x = 130 suggests 2^x = 5 and 8^x = 5^3, the latter of which can be rewritten 8^x = (2^3)^x = (2^x)^3 = 5^3, confirming that 2^x = 5.
Finally, 2^x = 5 => x = log5/log2, as stated on-screen.
When I got my first programmable calculator, an HP25, I wrote a simple bisection routine to solve the countless equations I came up with when taking physics exams. That is still how I would choose to solve problems like this. I leave math to the mathematicians.
Professors, who had been raised at a time when 4k of memory had to be rented because it was too expensive to buy, would tell me my approach was too expensive. I think about those comments every time I look at my stash of gigahertz laptops that are collecting dust.
The first thought is to replace 8 with 2^3, then cubic equation follows. We do not expect kids to remember Tartaglias formula so this is a dead end. If we hope for a integer solution it is good to remember that the free term (130) is a multiplicative of all 3 roots. Indeed it can be factired as 2*5*13. By checking al divisors of 130 (2,5,10, 13, 26, 65) it turns out that 5 is a solution and it is easy to see that it is also the only real solution.
If you like base-10 logs, log(5) = log(10/2) = log(10) - log(2) = 1-log(2), so you get x = 1/log(2) - 1. We know log(2) = 0.30103, so we can quickly compute it.
Before watching (so I might be very ashamed after!)... but it seems too easy. The BIG clue is that 8 is a power of 2 (2 cubed). So 8^x is (2^3)^x. And then we need a bit of substitution, so y for 2^x. Which means we have y^3 + y = 130 and it's clear there are very few options (as cubes get big quickly, so 6 is already a bust). It (y) must be 5. so 2^x = 5 and 8^x = 125. x is a bit more than 2 (2.2 or 2.3 ish). so i need a calculator to get exact number. or just express as logs as 2^x = 5 is same as x log 2 = log 5 therefore x = log 5/log 2. which is 2.322. And, of course, 8^2.322 + 2^2.322 is roughly 130. (130.019)
Just do prime factorials of 130 = 5^2 * 2 * 3
Once you have y^3 + y = 130 =
y^2 * (y+1) you now need to remember that 2 * 3 = 5 + 1 and you’re done
I’ll be honest with you I might be screwed I’ve done so many questions like these my first thought was “I’ll take the natural log of both sides” and my second thought was “maybe we can write in u as 3x my brain is so calculusified I can’t solve basic algebra
2:00 Just leave the 130 on the RHS and factorise the LHS:
y (y^2 + 1) = 130,
130 = 5 × 26 = 5 × (5^2 + 1)
y = 5.
Let y= 2^x
y^3 + y = 130
y = 5
x = log to the base 2 of 5.
There's a fair bit of luck in this approach but it genuinely took about 30s.
le indian solving this question preparing for jee 😎😎
Once you have a cubic, you have a known algebraic solution. Finding clever ways to extract the root quickly after this point is just a Sudoku puzzle.
At 3:30 factoring out y from y^3 -25y how does this become y(y^2-5) not y(y^2-25) ?
corrected 10 seconds later....ok.
Bro I don't know why but these question was asked by my teacher during maths quiz when I was in 9th stad. And out of 30 students 20 were able to solve these question within 2 mins I from India and these types of questions are very easy😂
I make this: 8^x+2^x=130
(2^x)^3+2^x=5^3+5
If we named 2^x as Y. Therefore, we have: Y^3+y=5^3+5
Which immediately means that Y=5
So, 2^x=5.
The equation show us that the answer is log 5 to the base of 2=x 😁
I got it right, that's good. But factored the cubic a little differently.
I love your videos, but I wonder why your audio signal is so low compared to most others?
You don’t have to say “equals to.” You can just say “equals.”
2::19 he says: "130 can be read as the product of 26 times 5" --- 130 = 26 x 5
--- WOW! UNBELIEVABLE! --- For a) 130 = 65 x 2 is the same as b) 130 = 13 x 10
*WHY* did he not choose a) y = 65y - 64y or b) y = 13y - 12y ???
I feel awful that logs were the one subject I never fully understood.
Me too. That ends our math careers!
I’m from India, an eleventh grade student and this is my soln-
8^x + 2^x = 130
2^3x + 2^x = 65 * 2
Take 2^x common
2^x (2^2x + 1) = 2 * 65
Now,
2^x = 2, (x=1)
And,
2^2x + 1 = 65
2^2x = 64
2^2x = 2^6, (x = 3)
Please correct me if I’m wrong
If you simply try putting x=3 into the original equation, you can clearly see that it is not correct, as it would equal 520, not 130.
Rewrite it, with a=2^x, a^3 + a = 130...guessing, lol, you can use a=5, that seems to work, 5^3 + 5 = 125 + 5 = 130...that means that 2^x = 5 is a solution, so that x= log_2(5), etc...
When u got y cube plus y eq 130 then it can be solve easily by value putting , so overall it became a very easy ques😅😅
This question is WAY too easy for either the maths olympiad or the Cambridge entrance exam. Both of those math tests are extremely difficult and you have to be of a rare kind to get anywhere with them!! This sort of question might be used at a Cambridge maths interview ... decent candiates would be expected to be able to solve this with ease and no prompting.
I’m just doing basic algebra with my son for his year 7 homework. Why isn’t it simplified to 10x = 130? So x = 13? That’s what my son has just learnt to do.
My regrets for not learning seriously on my high school days. I feel so stupid.
Use algebraic expression I know why you are using logarithmic expressions because of exponents but it is not necessary.
Where in the world, would you use this ability/facility in a practical sense?
I speak 5 languages and have degrees from Oxford and the Sorbonne and am a Fellow of the Royal Schools of Music. But my mind totally boggles at this, I am in awe. I guess you can’t have everything.
Too easy for me (JEE Aspirant) 😂 if Solving the cubic equation hit and trial is accepted
Maybe Cambridge High School?
8 base 10 = 100 base 2
=2^3
(2^3)^2= 2^3x
So 8^x +2 ^x = 130 = 2^3x +2^x = 130 base 10 = 1000001
Ln( (2^3x ) + 2^x ) = ln(1000001)
3x^2 = 2*ln(10^6)
3x^2 = 2 * 6
X= {(2^3)/√3}
8÷1.73 = x
X = 4.62
So it either had different matgatical definition of operations or tgey had no anwer because wrong hypotheses when
8^0.865361 + 2 ^0.865361 = 7.868223 base 2 = nearly exact 130 in base ten
So x is approximately 0.868223
So
the trick is 130=26*5
The answer is clearly Purple, because aliens don’t wear hats
Or we could just write it as: One hundred and dirty and solve it for y.
Xlog8 + xlog2 = log130
X (log 8 + log 2) = log 130
X =. Log 130 / (log 8 + log 2)
Problem solved?
9:14 Aug 4
This was NOT helpful. From the very moment when you split arbitrarily 130 into 26x5, the only thing that was clear is the fact that you knew the result beforehand, and nothing else. To anyone ignoring the result, the following steps were totally unclear. An demonstration of skills or insider knowledge is NOT an explanation. An explanation is a demonstration of how to proceed generally no matter what the outcome might be.
Sorry. Not good.
My approach was
8^x + 2^x = 130
2^3x + 2^x = 128 + 2
2^3x + 2^x = 2^7 + 2^1
3x + x = 7 + 1
4x = 8
x = 2
😂😂😂😂😂😂😂😂😂
simple, 8x15 =120 plus 2 x 5=10 so 130 😁
2^x = 5, then you put in log base 2 of x is 5
If you do not know math do not try to go there right? All they are about is math and nothing else?
8^x + 2^x = 130
2^3x + 2^x = 130
Let 2^x = y, then
y^3 + y = 130
y^3 + y - 130 = 0
y^3 - 25y + 26y - 130 = 0
y(y^2 - 25) + 26(y - 5) = 0
y(y^2 - 5^2) + 26(y - 5) = 0
y(y + 5)(y - 5) + 26(y - 5) = 0
(y - 5)[y(y+5) + 26] = 0
(y - 5)(y^2 + 5y + 26) = 0
For (y^2 + 5y + 26), d = b^2 - 4ac
So, 5^2 - 4*1*26 = 25 - 104 = -79
D is negative so y^2 + 5y + 26 has imaginary roots
Therefore y - 5 = 0
So, y = 5
2^x = 5
x * log(2) = log(5)
x = log(5)/log(2)
x = 0.698/0.301
Therefore, x ~ 2.32
Check :
8^2.32 + 2^2.32
=124.499 + 4.993
= 129.492
We took 3 places of decimal, as we go further the more accurate it get and at infinite places of decimal, it is 130
Yeah, I did it, but not elegantly like you.
Up to substitution, same.
Then I got y(y^2+1)-130=0,
concluded that the bracket cannot be negative, and thus y must be a positive number.
Then I tried numbers like a scrub and ended up with y=5, thus after resub, log2(5), and that was good enough for me lol
kinda easy ngl ((im lyinig)
130=125+5=5^3+5^1
can't stand when someone should pronounce it as 2 point 3 2 not 2 point thirty-two. you sound like a TV broadcaster who has failed in math.
Algebra Kyo lagaya re let 2^x=t and 8^x=t^3 by hit and trial 5 is a root .. therefore 2^x=5 so x= log base2 with 😂
at 3:27, why do you factorize y^3-25y into y(y^2- 5), it should be y(y^2-25), did I get something wrong?
sorry, I continued, and he corrected,
This is the easiest solution?
1. Change variable
2. Ruffini
3. Solve
Lol
Indian students laughing at the corner 😂.
Easy, seriously in india 6th graders solve these type of qs
@@tanvilalan3602 6th graders really ? You mean the average indian child learns this ? Or is it only in elite schools ?
X= 2.322 it took me 15 seconds to solve it.😂😂😂😂
While watching this video at 2 AM, my brain slowly dies...
this is brain nourishment
SAME!!! 2AM TOO!!
Same here😂
1:37 😢
I am watching at 5.37 with red eyes 🫥
I just brute forced this knowing that 8*8 is 64 and 8*8*8 is 512 it would have to be between 2 & 3 then via estimation and guess and check 2.32 came the closest in approximation. Even though I know this method is wrong on a standardized test with multiple questions and limited time. It is effective having a “feel” for numbers and proportions gets you through most of these tests
It is exactly how a physicist would calculate it
@@alonsobruni8131 **This is the way**
You're either into physics or computer science. Mathematicians look for the exact symbolic value.
Physics and chemistry…I’m algorithmically challenged 🤣
Used same method in about 2 minutes. Might have been able to do it analytical 40 years ago.. Hardly used any analytical math since poly technical. It's now a distant memory.
A middle schooler with a calculator: Nah Id win
I don't think you can do this with a calculator? whats the approach here
@@vikrantsingh6580you can actually with a sceintific calculator
not funny and you cant do it with calculator either
@@raheelkhan-ek9qe you actually can bruh
@@stannate how then bruh
From the line y^3+y-130=0
(y^3-125)+(y-5)=0
(y-5)(y^2+5y+25)+(y-5)=0
(y-5)(y^2+5y+26)=0
This way it took me 3 extra lines to get to the same result that the admin here took 6 lines.
At the second step, if you can see that 2^x = 5, you have the answer, and you'll have time to answer some other questions. These questions usually rely on having an insight like this, rather than on grinding out a solution, so you should be asking at this point, "is there a cube slightly less than 130?"
From 2^x=5 follows x=log5, base2. It is by a definition of logarithm!
that's what I thought!
@@orlevene9964 Not normally left like that. The day will come when you have to use a calculator without a base 2 log function, so it is best to know how to change to base 10 or natural log.
As soon as I saw that we had a cubic to solve, and 130 has a prime factorisation to 3 primes, 2x5x13, I knew one of them was going to be a candidate! Also if you dont do the 26-25 trick, and consider what the coeficients of general cubic would be, given there is no y^2 coeficient, you get the property that the sum of the roots is 0, and the product of the roots is 26.
I have never felt so bad at math in my life as I did trying to figure this problem out. Wow!!
2:25 i didint get it why you write y=26x25
yes also didn’t understand whats tge relation between 130=26*5 and 26y-25y?
I don’t fully understand it, but he uses this to substitute the 26y-25y for y (since 26-25=1), and then uses the factors of 130 to simplify the equation further
2:36 and 3:13
Hope this helps at least a bit 😅
I'm starting to freak out that after reading most of the comments I can't find any criticism of the line at 3:43 that claims
y^3 + 25y = y (y^2 - 5)
What am I missing here?
I had paused the video to look for comments. After continuing the video all is well, and I am no longer freaking out.
@@jonnamechange6854 Ha ha, so much anguish, happened to me too... and it took just a couple of seconds of patience to find salvation!
Yes - that gave me a WTF moment until he corrected himself a line or two later.
Cant we just take 26(y-5) on the other side and and divide the eq by y-5 whcih would give y²+5y+26= instantly 😅😅😅cbse 11th grader
Why doesn't using base 10 logs work here? 🤔
Such as:
x( log 8) + x(log 2) = log 130
Then solve for x 🤯
It does not matter what base you use: log5/log2 will always give the same number.
Can't we just do it like 2^3x+2^x=2⁷+2¹ but then we get two values for x =2.3 and x=1 So i guess this is wrong😅
It's comes to 8+2 = 10. X is 13. I failed algebra and did this in my head. 104 + 26.
Let us replace 2^x by X, then X^3 + X = 130, and evidently X = 5 is a solution.
As the function X -> X^3 + X is strictly increasing on R+, the value 130 is reached by the single X = 5.
So the unique real solution is given by 2^x = 5, i.e. x = Log 5 / Log 2. Thank you for your videos !
By inspection, 2 to the x = 5. Hence x = ln5/ln2
Is Cambridge this ez?
good job, but you have one minute left for the exam
sigh, as a mathematican, there are 3 solutions, log(5)/log(2) then log(-5/2+I*sqrt(79)/2)/log(2) and log(-5/2-I*sqrt(79)/2)/log(2) not one.
Super 👍 from Algeria
I watch it for sleep works great
ln5/ln2 is the ans. I am currently in 11th and i did this by basic algebra.
Assume 2^x = t and then youll get a cubic eqⁿ. One of its roots by inspection is 5, and the other remaining Quad. Eqⁿ is a always +ve factor, so it doesnt matter. Hence there is only 1 solⁿ, y=5
=> 2^x=5, ln(2^x)= ln5, x(ln2)=ln5
x=> ln5/ln2 or log5 base 2👍🏻
You don't need natural logs. Log base 10 works perfectly as does any base. ln isn't wrong but you don't need it.
My approach before Watching this video
put 2^x=t
Now, we have a simple cubic equation
t³+t=130
=>t(t²+1)=130
Now,by trail and error method
Factors of 130 = 2×13×5
Putting t=5 in LHS,we get 5(25+1)=5×26=130
Hence the value of t=5
As,2^x=t (Taking logarithm both sides)
=>xln2=lnt
Put the value of t=5
x=ln5/ln2
Got the answer by simple algebra manipulation and took 2 minutes to write and seconds to solve
It is much easier to follow also :)
Once you have the cubic in t, just try looking for a perfect cube that's not much less than 130.
Trying 125 = 5³, we have the solution, because 5³ + 5 = 130. The rest follows as you describe.
Fred
Yup you just need basic 10th grade mathematics fundamental concepts and you can solve it easily within 3 minutes This made me realise how hard Jee Advance is.
I'm impressed to see so many alternative solutions. I see that the followers of this channel are true mathematicians. Well done! I would have solved the cubic equation using synthetic division. Best regards from Bogota.
Fun fact: 130 = 2⁷ + 2
Very easy as a jee aspirant
I agree 💯
Just write “black lives matter” 1000 times
Perfect! You are ready for NASA!
Newton’s method gives an approximation of log5/log2 by nine significant digits. Super cool problem!
В который раз вижу ролик этого автора и никогда задача не ставится корректно. Найдём корни - КАКИЕ? - действительные, комплексные, рациональные, гиперкомплексные😅, ещё какие-нибудь... Это, извините, задача вступительных экзаменов, тут надо точно формулировать задание. А то в одном месте ищет ВСЕ комплексные решения, а тут - "извините, комплексные слишком сложные, мы их искать не будем". Ха-ха. Это не серьезный подход
I forgot how to "correctly" do this, but I realized 8 is the cube of 2 so i just did 2x2x2xetc until I got to 128 and then I saw I just needed to add 2. So 8^(1/3) + 2^7...
And then I remembered both are x so they have to be the same number and I'm just gonna go to bed.
What if i applied ln (2^x) instead of log(2^x)? in the second last step?
It you use a calculator they will come out to be the same.
The reason being log or ln values are constants and dividing will yield the same ratio, i.e., ln5/ln2 = log5/log2.
Another way to reason this would be to use the "change of base property" of logarithms.
Log5(base10) ÷ Log2(base10)
=
(Ln5/Ln10) ÷ Ln2/Ln10)
Here Ln10 cancels out and we are left with "Ln5/Ln2"
Thus Ln5/Ln2 = Log5/Log2
***Hope this helps😊***
Oh okay I see you went for the longer method. The moment I saw the question the intuition was very clear because of exponential I immediately used log on both sides, and put the values to the calculator.
Using log on the left hand side does not help you in any way for solving this equation.
This is too simple to be an olympiad problem. Just take 2^x to be a variable 'a'. Now solve the quadratic and find the value of x
First thought: "Oh, that's simple, just 128 + 2, which is 2⁷ + 2¹"
Then after diving in: "Something between 2 and 3, if they chose =68 or =520 this would be simple"
Using bracketing (artillery method for finding range of target) and trial and error method should be able to do it in one or two minutes which would be much faster and more importantly less prone to algebraic error, but of course less elegant. Also if it is a multiple choice test you could quickly try the different answers and find the correct one in very little time.
Miss those jee advanced days. Glad to see our students are much advanced, if this is the standard of the paper
How come no Nobel Prize winners or Fields Medallists has ever come out of any IIT ?
There should be 3 complex roots if the principal value of ln z is used. One of which is the solution presented.
8^x + 2^x = 130. 8 = 2^3 so we have 2^3x + 2^x = 130. You can say, immediately, that by inspection 2^x = 5 because 5^3 + 5 = 125 + 5 =130. Thus x = log5/log2 . I can't do that in my head - I know it's 2 point something and my best guess would be around 2.3 - but a few seconds on a calculator tells me that x = 2.3219 to 5 sig fig. Or if you need a little more time you can try 2^x = y so y^3 + y = 130 and again you can immediately see that y = 5. The rest follows - no need for all that manipulation for a simple problem which is what, I think, the Cambridge examiners would want to see. No need to use a sledgehammer to crack a peanut.