🤩 Hello everyone, I'm very excited to bring you a new channel (SyberMath Shorts) and my first video in short form! Enjoy...and thank you for your support!!! 🧡🥰🎉🥳🧡 www.youtube.com/@SyberMathShorts
You make the problem so hard....they have the same power so we can subtract the whole number 5-3...and then just copy the exponent x...then the result is 2 to the x power...to get 16, of course, x is equal to 4...2 to the fourth power equals 16
5^x - 3^x = 16 This will have only 1 solution because as x increases, 5^x - 3^x increases monotonically. 5^x = 3^x + 4^2 This is the 3, 4, 5, right triangle (Pythagoras) with x = 2
here it works because right term is positive , but 5^x - 3^x is only monotonic for x>0 If the problem was 5^x - 3^x = -0,1 for example, there are 2 solutions, (numerically about) -1,5446 and -0,2894 Derivate of (5^x - 3^x) is null for x= - { ln [ ln(5)/ln(3) ] / ln(5/3) } = -0,7475 and the minimum of (5^x - 3^x) is then (numerically about) -0,1396 That means that 5^x - 3^x = c has no solution for c < -0,1396 has 2 solutions for -0,1396 < c < 0 has 1 solution for c > 0
@@rickdesper The structure is highly ramified. It's quite interesting to do a contour plot equating real and imaginary parts in something like Mathematica.
@@khronosolympian908 I will prove that there is only one solution by means of two demonstrations following: Demonstration 1: Doing x = a results 5ª - 3ª = 16 5ª - 3ª = 4² 3ª + 4² = 5ª From the Pythagoras Theorem we have 3² + 4² = 5² Being a = 2, then x = 2. ============= Demonstration 2: For x = 0: 5⁰ - 3⁰ = 0 For x = 1: 5¹ - 3¹ = 2 For x = 2: 5² - 3² = 16 Therefore, x = 2.
@@juliocesarsilvapontes3717 No, you just showed that 2 is one of the solutions, without proving that it is the unique solution of the equation... And your second demonstration is not even a demonstration, you just tried 0, 1 and 2. I'm starting to think that you're just a troll x)
When you write the equation as : 5^x = 3^x + 4^2 you immediately have to think about the first Pythagore's triplet (3,4,5) which gives you 3^2+4^2 = 5^2 . Therefore x=2 is the only solution for this triplet.
As soon as I saw this problem, I thought of a right triangle of (a=3, b=4, c=5) Therefore, it was very easy to think that x=2. However, it was not easy to find whether there was another x except x=2. I think other solutions may exist in the complex number domain.
I can't see any other possible solution. the difference just gets greater each time you multiply by 5. But if there is another possibility, I would like to understand it
The problem with that is by assuming Pythagorean theorem applies here, you're already assuming the answer is two. So saying the answer is two is a tautology.
I find a quick and nongraphical solution: Let a=4 so (a+1)^x - (a-1)^x = a^2 Apply a²-b²=(a+b)(a-b) to equation. (2a)^(x/2) • 2^(x/2) = a^2 (4a)^(x/2) = a^2 Replace a to 4 16^(x/2) = 16 x/2 = 1 x = 2
We can make another attempt to find the same solution without considering graphical representation. Now we can write the equation as (5^(x/2))^2 - (3^(x/2))^2 = 16. This is difference between square of two terms. Therefore, it can be written as ((5^(x/2)+ (3^(x/2))(((5^(x/2) - (3^(x/2)) = 16. Let ((5^(x/2)+ (3^(x/2)) = a and (((5^(x/2) - (3^(x/2)) = b so that ab = 16 and a > b as x>0. Then we have only two pair of solutions for a and b. (i) a=16 and b = 1 and (ii) a = 8 and b = 2. The first set on solving will result 5^(x/2) = (16+1)/2) i.e, x = 2.66 and 3^(x/2) = (16-1)/2 i.e., x = 3.66 both of which are not unique. Therefore it is not a solution. Solving 2nd set where a = 8 and b = 2, we get 5^(x/2) = (8+2)/2 = 5 so that x = 2, again 3^(x/2) = (8-2)/2 = 3, so that x = 2. Both the values of x are now same. So the solution is x = 2.
In case, if we have a problem say 5^x-3^x=2, then how can we solve to get x=1 as solution. Can the product of sums and difference of two numbers can it give the solution ?
We could easily find the fact that 5^x - 3^x is monotonous function and difference is greater 0 with x ≥ 0 and less than zero with x < 0. Considering this clause we find out that 5^x - 3^x function intersects y axis with value 16 in one only point - (2;16)
For those who are throwing their their knowledge in comment section, let me clarify you that there is a possibility that there exist one or more answers rather than 2 and those answers may be in complex numbers So to find out if those complex numeral answer exists or not, we have to solve this question by the method shown in this video
I approached it slightly differently. If you define f(x) = 5^x - 3^x, then f(x) is a strictly increasing function and hence injective. Therefore if a solution exists to f(x)=c then x must be unique. We can see that f(2) = 5^2-3^2 =16. Hence x=2 is the only solution.
I used the same approach but first I noticed that there can't be any solution 5^x-3^x For x>1, f'(x)=x(5^(x-1)-3^(x-1))>2x>0 then f is strictly rising for these values of x. Since f(1)=2 and f(x)---->+inf when x--->+inf and since f is a continuous function, the theorem of intermediate values says us that there is a value where f(x)=16 and this value is unique since f is strictly rising. But f(2)=5²-3²=16. Then 2 is the only solution.
If we called 16 a^x, where a is a constant, then we have the equation in the form a^x = b^x + c^x. According to Fermat's last theorem, we know there are no integer solutions beyond x = 2
Express the equation modulo 3 and apply Fermat's Little Theorem one finds x is even. Then the left hand side factors as the sum of squares. The sum of those factors must be twice a power of five. Only two factors of 16 sum to twice a power of five, 2 and 8. That sum is ten so the power is one. This power is half x, so x is 2.
Haha - I guessed 2 as the answer very quickly, but was unsure how to prove my answer. You went thru a lot of hoops and finally used the "trial-and-error" method that I used initially. This reminded me to the Pythagorean Theorem for a right triangle using 3-4 as the sides and 5 as the hypotenuse. As soon as I subtracted 3^x from both sides, it was obvious that the answer was 2. 5^2 = 3^2 + 16.
The problem with that is by assuming Pythagorean theorem applies here, you're already assuming the answer is two. So saying the answer is two is a tautology.
@@jessekwb5035 Having been a math major 50 years ago, this answer just jumped off the page at me. Not sure about the tautology thing, as that would just introduce a discussion. Math is simple - one answer per problem (save binomial equations and square roots). There is never discussion about math. If you get the ONE answer, then nobody can argue that (unless you're from another dimension).
@@ClemsonTiger75 First off, there's not always just one answer per question, a math major should know that. Second, trial and error simply doesn't seem like a viable strategy to me, it works with low and simple numbers like this, but that doesn't mean it'll always work. And math isn't just about the answer, proof is important. And assuming the answer to be true isn't the best proof. Sure, you can now work backwards and that would be good proof, but when you're guessing, it's not always going to work out.
We can solve with algebra , remember hasil pengurangan angka kuadrat = hasil perkalian penjumlahan angka angkanya dengan selisih angka angkanya ,dgn demikian kita misalkan 5^2-3^2=(5+3)(5-3)=8x2=16, sehingga nilai x=2, secara kebetulan terbukti
What i do in secondary school, in objective questions for this kind of question since i cannot stress my self. I would pick among the options, and fit it in the question, if it gives 16, that's the answer. So in this question since i didn't see any option, i just started with x = 2 5² - 3² = 25-9= 16, wow i got it the deceptive way 😳🤣😜😜😜
Consider both equations in mod 3. The left equation is (-1)^x. The right equation is 1. Therefore x must be even number, and x=2m below. 5^2m-3^2m=(5^m+3^m)(5^m-3^m)=16 From 5^m+3^m>5^m-3^m 5^m+3^m=16,8 5^m-3^m=1,2 Therefore only m=1 is suitable, and from x=2m the answer is x=2.
Solution to this equation is so simple and at our fingertips. But showing that it is the unique solution, you did a good job! Interesting! Post more of this kind please!
If you know your pythagorean triplets, it's easy to see x=2. Other than that, rewrite 16=2^4 and use Bertrand's Postulate to show that there are no other solutions for when x>2.
Левая часть уравнения - монотонно возрастающая функция. Правая- постоянная величина. Значит, уравнение имеет не более одного корня. Подбором находим, что х= 2.
X=2 is only answer From pyrhagorous theorem AC^2-AB^2=BC^2 AC=√(5^x), AB=√(3^x), BC=√(16) (√(5^x) )^2-(√(3^x) )^2=(√16)^2 (5^x) )-((3^x) )=(16) Take R. H. S. =(5^x) - (3^x) =(5^2) -(3^2) =(25) -(9) =16 =L.H.S.
Lol I just put 2 in n got it, thanks though for calc class reminder cause going back to physics. I’m 41, going back to school n trying to refresh my memory :)
The problem with that is by assuming Pythagorean theorem applies here, you're already assuming the answer is two. So saying the answer is two is a tautology.
If you know from basic geometry that 3^2+4^2=5^2 then just change 4^2 to 16 and move -3^2 to the right side of equation. Or just guess and check because you know, or should know, it has to be small integer. But my way is more fun
It is very interesting to solve the problem via the # of intersections of 2 functions to determine # of solutions, and then determine these small # of solutions by observation or trial-and-error. However, in this case there is a easier direct non-trial-error way using basic algebra (btw, a similar approach was used in another video on youtube for a slightly easier problem w/ a similar structure; so I cannot lay claim to this approach, but the simple extension--which I'll mention when we come to it---is mine. Here is the solution: 5^x - 3^x = 16 ==> (5^{x/2})^2 - (3^{x/2})^2 =16. Let a = 5^{x/2} and b = 3^{x/2}, and note a, b are > 0. Thus we have a^2 - b^2 = 16 ==> (a+b)(a-b) = 16 = 4*4 = 8*2 (the consideration of multiple 2-number factors is mine when the RHS has more then 2 prime factors---the aforementioned video had exactly 2 prime factors--, and we solve the problem for each of these **valid** factors [those w/ unequal numbers] to see which yields integer solutions) Since a+b > a - b, the only valid factor is 8*2 (in general, there may be more than one valid 2-number factor and we need to try each for possibly multiple integer solutions for x). Thus (a+b)(a-b) = 8*2 ==> a+b = 8, a-b = 2 ==> 2a = 10 ==> a = 5 ==> 5^{x/2} = 5 ==> x/2 = 1 ==> x = 2 as the only integer solution (a verification---a given--by plugging in x=2 in the original equation, proves it).
Brilliant this is another way of looking at math problems like another idea it could be use I some problems so as many ideas like this we have somewhere in our head we would be better at solving problems in general
Fast analytic method after guessed x=2: Let f(x) = 5^x - 3^x for x < 0, we don't need to worry too much because 5^x < 3^x < 1, which means f(x) must be negative. f(0) = 0 f'(x) = 5^x * ln(5) - 3^x * ln(3) which is always > 0 when x >= 0 So f(x) is strictly increasing when x >= 0 x=2 is the only solution.
Other than by Inspection - in a few seconds - or trial and error - in a few seconds, you could write this as: 5^x - 3^x = 16 = 4^2 (or option 2 as 2^4). Let's try the first option as x=2 then the seconds as x =4. So rearrange and you get 5^x = 3^x +4^2 (implying x = 2) or 5^x = 3^x +2^4 (implying x = 4), now 5^2 = 3^2 + 4^2 (25 = 9 + 16) but 5^4 3^4 + 2^4 (as 625 81 + 16). So if you just guessed x= 2 - or you knew a 3,4, 5, right angle triangle - the answer would fall out immediately. You could also attempt to solve this for 16 = 2^4, but if you try for x = 4 you would see it doesn't work as 5^4 3^4 + 2^4 so x must equal 2 is the only answer.
I don't think this is really the correct way to solve this equation - it's just one way of brute force - not purely mathematical. I think it is better to consider that 5 to the same power (except zero) always has 5 as the last digit - because xx5 * 5 = xxx5 etc. On the other hand 3^x must have the last digit 9 to give the number with last digit 6 on the right! And that's it! Now we only have to prove when 3 ^ x gives the number with the last digit 9? And this only happens with the pattern: 3^(2 + 4 * n). And because 5^(2 + 4 * n) - 3^(2 + 4 * n) is an increasing function > 16 for n > = 1 than n must be 0. So this is the correct solution => 5^(2+0) - 3^(2+0) = 16 => x = 2!
Although the process did not seem rigid but it is well explained. I am quite interested to know that such a "difficult" problem can be solved by trial and error !
Yes, there are, x=2+2kπi is a solution for each integer k. Setting y=e^x, the equation gets y^(ln 5)-y^(ln 3)=16. It is easy to see that y will be the same if x changes over 2πi. We can go a bit further. Let x=a+bi. Then 5^x and 3^x are somewhere on the circles with radii 5^a and 3^a, respectively. The phase difference depends on the imaginary part. We could try to solve the system 5^a cos(b ln(5)) - 3^a (cos b ln(3)) = 16 5^a sin(b ln(5)) - 3^a (sin b ln(3)) = 0 ... but tbh I don't really expect to find any x with real part other than 2.
I just noticed the equation is: 5^x = 3^x + 4^2, and the common 3x4x5 triangle works to solve this via the Pythagoras theorem when x=2. Also, it appears to only have 1 answer, because 5^x is increasing and 3^x is decreasing (but less than 5^x) and 4^2 is a constant. Hence, the only answer is x=2. No other maths required. boom. answer in like 5sec.
My preferred method is to change the equation to ((5^(x/2))^2)-((3^(x/2))^2)=8×2, then substitute, respectively, u and v for 5^(x/2) and 3^(x/2). We then have (u^2)-(v^2)=8×2. This factors to (u+v)(u-v)=8×2, which makes for a relatively easy equation system of u+v=8 and u-v=2. U and v are, respectively, 5 and 3, which are easy to solve for each exponent of x/2. I don't like guess and check. I like definite general rules for a given equation form of any type.
Ça vient direct pour ceux qui maîtrisent le théorème de Pythagore concernant le triangle rectangle avec angle droit a²+b² = c² A partir de là la solution se voit 16 étant le carré de 4 on ne fait qu'appliquer le théorème de Pythagore
let assume original equation is 5^x-3^x=a chose a number for x then put in right side of this equation that i got from original equation x=ln(a-3^x)/ln5 and get new X and repeat those steps, after 3 or 4 steps you get approximate answer. in this way it doesn't mater a is equal 16 or any number
It would be nice to see this used to solve for any given x. Solving this for x in terms of y would be very cool. You can eyeball this and see that it's 2, but isolating x and solving for it for any given y would be nice.
5^x - 3^x = 4^2. (3,4,5) are known pythagorian numbers, ergo x = 2. Done. This is beyond easy. Also it is known that the exponential function is monotonous, meaning there won't be any other value besides x = 2 that fulfills the same role as x = 2
This is a simple equation, just write 5=3+2 and then apply binomial Newton, then remove 3^x. Simplify by 2 from both sides, then the remaining sum should be 8. The only case is ×=2. Since the formula increases with x.
When ever I saw that question I substituted x=2 I got the answer and fast forwarded the video to verify my answer. I got correct answer that is x=2 and I got my answer in less than 5 seconds.I know many of got the answer like me.
😂😂😂😂 bro you made this problem complicated 😂😂😂😂 idk why cuz you can use this method instead Here it is :- First write 5^x -3^x = 16 Now write 16 as 5²-3² After comparing the base we get x =2 easyyyyyy
👍I guessed the solution at the starting of video. But finding how much possible solutions this equation could have was important part, so that we can guess and find all possible solutions. Good tutorial, keep it up.👍
🤩 Hello everyone, I'm very excited to bring you a new channel (SyberMath Shorts) and my first video in short form! Enjoy...and thank you for your support!!! 🧡🥰🎉🥳🧡
www.youtube.com/@SyberMathShorts
Hi.
I tried “x=1” and saw that was wrong. Then I tried “x=2” and saw that was correct. That’s how engineers do it
You make the problem so hard....they have the same power so we can subtract the whole number 5-3...and then just copy the exponent x...then the result is 2 to the x power...to get 16, of course, x is equal to 4...2 to the fourth power equals 16
Yeah, this video could have been five minutes shorter.
@@emotional7626 That is not correct, although I suspect you know that. :-)
Put 2 in power and we will get 16 both sides
@@emotional7626 thats wrong
5^x - 3^x = 16
This will have only 1 solution because as x increases, 5^x - 3^x increases monotonically.
5^x = 3^x + 4^2
This is the 3, 4, 5, right triangle (Pythagoras)
with x = 2
Nice!
here it works because right term is positive , but 5^x - 3^x is only monotonic for x>0
If the problem was 5^x - 3^x = -0,1 for example, there are 2 solutions, (numerically about) -1,5446 and -0,2894
Derivate of (5^x - 3^x) is null for x= - { ln [ ln(5)/ln(3) ] / ln(5/3) } = -0,7475 and the minimum of (5^x - 3^x) is then (numerically about) -0,1396
That means that 5^x - 3^x = c
has no solution for c < -0,1396
has 2 solutions for -0,1396 < c < 0
has 1 solution for c > 0
You did not explain why 5^x - 3^x increases monotonically with x (which is wrong for x < -0.7475).
Wow, I never think it
Very good logic
I saw x=2 immediately. The limit argument sealed it as the single solution. Thank you and well done!
No problem! Thank you! 😊
The 'argument' was an elaborate ruse...a 'Rube Goldberg" math trick.
If it was mcq I would have guessed that only
其實,通常一個正常既學生經常用(畢氏定理)呢個方式去計算"直角三角形"既長度都會好快知道答案.
Please can you explain how can i know the number of solutions using a limit ?
There is an infinite family of solutions in the complex plane. An example is the number that is close to 1.604 + 3.684 i.
Since exp(2 pi i) = 1, this is true of most exponential equations. Every function b^x with real, positive base b is cyclic on the imaginary axis.
@@rickdesper The structure is highly ramified. It's quite interesting to do a contour plot equating real and imaginary parts in something like Mathematica.
Intuition is the highest form of knowledge. I knew the answer even before finishing reading the problem.
Solution
5ª - 3ª = 16
5ª - 3ª = 4²
3ª + 4² = 5ª
From the Pythagoras Theorem we have
3² + 4² = 5²
Therefore:
a = 2 (or x = 2).
Why would it be the only solution?
@@khronosolympian908 Because there is only one value of x that substituted into the exponential expression gives 16.
@@juliocesarsilvapontes3717 How do you prove that?
@@khronosolympian908 I will prove that there is only one solution by means of two demonstrations following:
Demonstration 1:
Doing x = a results
5ª - 3ª = 16
5ª - 3ª = 4²
3ª + 4² = 5ª
From the Pythagoras Theorem we have
3² + 4² = 5²
Being a = 2, then x = 2.
=============
Demonstration 2:
For x = 0:
5⁰ - 3⁰ = 0
For x = 1:
5¹ - 3¹ = 2
For x = 2:
5² - 3² = 16
Therefore, x = 2.
@@juliocesarsilvapontes3717 No, you just showed that 2 is one of the solutions, without proving that it is the unique solution of the equation... And your second demonstration is not even a demonstration, you just tried 0, 1 and 2. I'm starting to think that you're just a troll x)
When you write the equation as : 5^x = 3^x + 4^2 you immediately have to think about the first Pythagore's triplet (3,4,5) which gives you 3^2+4^2 = 5^2 . Therefore x=2 is the only solution for this triplet.
My first thought as soon as I saw this
Same bro, literally the first thought everyone (atleast knowledge of 9th class maths) would get.
Exactly 💯
Why does your argument imply that x=2 is the only solution?
Exactly
As soon as I saw this problem, I thought of a right triangle of (a=3, b=4, c=5)
Therefore, it was very easy to think that x=2.
However, it was not easy to find whether there was another x except x=2.
I think other solutions may exist in the complex number domain.
I can't see any other possible solution. the difference just gets greater each time you multiply by 5. But if there is another possibility, I would like to understand it
@@SteveWhiteDallas he is talking about complex numbers bro
I also thought in terms of a 3, 4, 5 right triangle.
It is a right angled triangle with sides 3, 4 and 5. Pythagoras's Theorem: the longest side squared = the sum of the other two sides squared.
Pythagorean
a^2 + b^2 = c^2 where c is the hypotenuse and a and b are some other stuff
The problem with that is by assuming Pythagorean theorem applies here, you're already assuming the answer is two. So saying the answer is two is a tautology.
That's not what the strawman said! 😂
I find a quick and nongraphical solution:
Let a=4 so (a+1)^x - (a-1)^x = a^2
Apply a²-b²=(a+b)(a-b) to equation.
(2a)^(x/2) • 2^(x/2) = a^2
(4a)^(x/2) = a^2
Replace a to 4
16^(x/2) = 16
x/2 = 1
x = 2
This is the most correct solution, i encountered so far! cheers bro..
You are a genius 😯😯
You made such a big logical jump that it became really hard to understand
We can make another attempt to find the same solution without considering graphical representation. Now we can write the equation as (5^(x/2))^2 - (3^(x/2))^2 = 16. This is difference between square of two terms. Therefore, it can be written as ((5^(x/2)+ (3^(x/2))(((5^(x/2) - (3^(x/2)) = 16. Let ((5^(x/2)+ (3^(x/2)) = a and (((5^(x/2) - (3^(x/2)) = b so that ab = 16 and a > b as x>0. Then we have only two pair of solutions for a and b. (i) a=16 and b = 1 and (ii) a = 8 and b = 2. The first set on solving will result 5^(x/2) = (16+1)/2) i.e, x = 2.66 and 3^(x/2) = (16-1)/2 i.e., x = 3.66 both of which are not unique. Therefore it is not a solution. Solving 2nd set where a = 8 and b = 2, we get 5^(x/2) = (8+2)/2 = 5 so that x = 2, again 3^(x/2) = (8-2)/2 = 3, so that x = 2. Both the values of x are now same. So the solution is x = 2.
why a, b take integral values ?
why did you put it as X/2 if in the question it's showing 5^x and 3^x ? where did you get that?
@@aj14rebo Just to make it as the difference between two squares so that we can use them as the product of two terms such as (a + b)(a - b) forms.
The best solution I have seen thus far.👍🏽
In case, if we have a problem say 5^x-3^x=2, then how can we solve to get x=1 as solution. Can the product of sums and difference of two numbers can it give the solution ?
Hey, actually on observing the equation carefully we get to know that this is forming a Pythagoras equation with a famous triplet 3,4,5.
Yes!
String theory
Yes
Yep, it took a moment to realize that those bases look familiar and 16 can be rewritten.
No sherlock
We could easily find the fact that 5^x - 3^x is monotonous function and difference is greater 0 with x ≥ 0 and less than zero with x < 0. Considering this clause we find out that 5^x - 3^x function intersects y axis with value 16 in one only point - (2;16)
Good thinking!
Did the same :)
I am 75 yo, and have a love for math. You are helping to keep my mind sharp.
Glad to hear that! Good job for you! 🥳🎉🤩
For those who are throwing their their knowledge in comment section, let me clarify you that there is a possibility that there exist one or more answers rather than 2 and those answers may be in complex numbers
So to find out if those complex numeral answer exists or not, we have to solve this question by the method shown in this video
Nerd
Here is a more interesting equation to solve. 3^x - 5^x = 1/15 Do not expect a pretty solution!
Sure ... there are two solutions : solved numerically => about -2,0787 and -0,1626
@@tontonbeber4555 Yes. Also, it turns out that 3^x - 5^x can be approximated by
-0.5 * x * exp(1.3 * x) for x
I approached it slightly differently. If you define f(x) = 5^x - 3^x, then f(x) is a strictly increasing function and hence injective. Therefore if a solution exists to f(x)=c then x must be unique. We can see that f(2) = 5^2-3^2 =16. Hence x=2 is the only solution.
I used the same approach but first I noticed that there can't be any solution 5^x-3^x
For x>1, f'(x)=x(5^(x-1)-3^(x-1))>2x>0 then f is strictly rising for these values of x.
Since f(1)=2 and f(x)---->+inf when x--->+inf and since f is a continuous function, the theorem of intermediate values says us that there is a value where f(x)=16 and this value is unique since f is strictly rising.
But f(2)=5²-3²=16.
Then 2 is the only solution.
i like that approach
5 square - 3 square =16
(5X5)-(3X3)=16
25-9=16
9th grader right here
i took a completely different approach with wrong and careless calculations, but somehow got the right answer lol
That was a very long winded way of saying "Just guess"
If we called 16 a^x, where a is a constant, then we have the equation in the form a^x = b^x + c^x. According to Fermat's last theorem, we know there are no integer solutions beyond x = 2
Interesting!
Watch the latest video of wishing happy new year in the language of Mathematics
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2^3+2^3=16
フェルマーの最終定理は自然数しか扱えない
Who said anything about integer solutions? How about complex numbers with real coefficients? Would like to see that.
Express the equation modulo 3 and apply Fermat's Little Theorem one finds x is even. Then the left hand side factors as the sum of squares. The sum of those factors must be twice a power of five. Only two factors of 16 sum to twice a power of five, 2 and 8. That sum is ten so the power is one. This power is half x, so x is 2.
nice one!
interesting
Haha - I guessed 2 as the answer very quickly, but was unsure how to prove my answer. You went thru a lot of hoops and finally used the "trial-and-error" method that I used initially. This reminded me to the Pythagorean Theorem for a right triangle using 3-4 as the sides and 5 as the hypotenuse. As soon as I subtracted 3^x from both sides, it was obvious that the answer was 2. 5^2 = 3^2 + 16.
i think we can solve this by applying logarithm.
The problem with that is by assuming Pythagorean theorem applies here, you're already assuming the answer is two. So saying the answer is two is a tautology.
@@jessekwb5035 Having been a math major 50 years ago, this answer just jumped off the page at me. Not sure about the tautology thing, as that would just introduce a discussion. Math is simple - one answer per problem (save binomial equations and square roots). There is never discussion about math. If you get the ONE answer, then nobody can argue that (unless you're from another dimension).
@@ClemsonTiger75 First off, there's not always just one answer per question, a math major should know that. Second, trial and error simply doesn't seem like a viable strategy to me, it works with low and simple numbers like this, but that doesn't mean it'll always work. And math isn't just about the answer, proof is important. And assuming the answer to be true isn't the best proof. Sure, you can now work backwards and that would be good proof, but when you're guessing, it's not always going to work out.
We can solve with algebra , remember hasil pengurangan angka kuadrat = hasil perkalian penjumlahan angka angkanya dengan selisih angka angkanya ,dgn demikian kita misalkan 5^2-3^2=(5+3)(5-3)=8x2=16, sehingga nilai x=2, secara kebetulan terbukti
Useful and amazing as usual, thx teacher 😊
What i do in secondary school, in objective questions for this kind of question since i cannot stress my self.
I would pick among the options, and fit it in the question, if it gives 16, that's the answer. So in this question since i didn't see any option, i just started with x = 2
5² - 3² = 25-9= 16, wow i got it the deceptive way 😳🤣😜😜😜
我有同感，在視頻中只知道函數的遞減性，我以為要用夾擠的關係證明答案，我英聽不大好，也許有什麼我沒聽出來
Consider both equations in mod 3.
The left equation is (-1)^x.
The right equation is 1.
Therefore x must be even number, and x=2m below.
5^2m-3^2m=(5^m+3^m)(5^m-3^m)=16
From 5^m+3^m>5^m-3^m
5^m+3^m=16,8
5^m-3^m=1,2
Therefore only m=1 is suitable, and from x=2m the answer is x=2.
Solution to this equation is so simple and at our fingertips. But showing that it is the unique solution, you did a good job! Interesting! Post more of this kind please!
Glad it helped!
The solution, x=2, is so instantly obvious I thought there must be a twist coming in the video.
If you know your pythagorean triplets, it's easy to see x=2. Other than that, rewrite 16=2^4 and use Bertrand's Postulate to show that there are no other solutions for when x>2.
I solved it 10 seconds by trial and error method
I loved the explainings and the resolution !!!!!!!!!
😍😍😍😍😍😍
💪🤙👌
👏👏👏👏👏👏
Thank you!!! 💖😍😊
Левая часть уравнения - монотонно возрастающая функция. Правая- постоянная величина. Значит, уравнение имеет не более одного корня. Подбором находим, что х= 2.
ですね！
x=2⇒5^x-3^x=16
あなたの議論から、解はただ一つであるからx=2で必要十分。
@@sibuyayuto3289 а теперь нужно доказать, что нет других решений.
@@ЗояШаромет
absolutely
X=2 is only answer
From pyrhagorous theorem
AC^2-AB^2=BC^2
AC=√(5^x), AB=√(3^x), BC=√(16)
(√(5^x) )^2-(√(3^x) )^2=(√16)^2
(5^x) )-((3^x) )=(16)
Take R. H. S.
=(5^x) - (3^x)
=(5^2) -(3^2)
=(25) -(9)
=16
=L.H.S.
I first saw the q, and immediately thought ,16 = 4^2
3^2 + 4^2 = 5^2
thats how 8th grade students do it
A very simple exponential equation. By observation, it looked likely X was 2. By mental calculation thus it was so confirmed in 6 seconds.
Should’ve taken 1 second. This isn’t even binary
@@AYVYN Hey, cut me some slack, I'm past my Sixties and had a few wines :-)
Great video! Keep up the good work 👍🏻
Thank you! 👍
Good method. Another method is working in mod 4 or 5 we see that x is even. Then we can factor and solve
Great 👍
Lol I just put 2 in n got it, thanks though for calc class reminder cause going back to physics. I’m 41, going back to school n trying to refresh my memory :)
Yes. I recalled the right-angled triangle with sides 3, 4, and 5. Hence x=2 is a solution.
The problem with that is by assuming Pythagorean theorem applies here, you're already assuming the answer is two. So saying the answer is two is a tautology.
This can be easily solve by this 5^x-3^x=16=5^2-3^2 so base is same then x=2
I love the way you approach the exercice. Great job ! 🙏🙏🙏
Thank you!
Solution
For x = 0:
5⁰ - 3⁰ = 0
For x = 1:
5¹ - 3¹ = 2
For x = 2:
5² - 3² = 16
Therefore, x = 2.
And for higher powers, the difference will be larger
You deserve so much better!
If you know from basic geometry that 3^2+4^2=5^2 then just change 4^2 to 16 and move -3^2 to the right side of equation.
Or just guess and check because you know, or should know, it has to be small integer.
But my way is more fun
It would be much easier if you consider two functions:
f(x)=5^x
g(x)=16+3^3
X=2
Obviously, I enjoyed it
I love the way you explained it
Geo drama.
This is a simple pythagorean triplet:3,4,5. x=2
but I see many saw that too.
we have all had that triplet pounded into our heads.
It is very interesting to solve the problem via the # of intersections of 2 functions to determine # of solutions, and then determine these small # of solutions by observation or trial-and-error. However, in this case there is a easier direct non-trial-error way using basic algebra (btw, a similar approach was used in another video on youtube for a slightly easier problem w/ a similar structure; so I cannot lay claim to this approach, but the simple extension--which I'll mention when we come to it---is mine. Here is the solution:
5^x - 3^x = 16 ==> (5^{x/2})^2 - (3^{x/2})^2 =16.
Let a = 5^{x/2} and b = 3^{x/2}, and note a, b are > 0.
Thus we have a^2 - b^2 = 16 ==> (a+b)(a-b) = 16 = 4*4 = 8*2 (the consideration of multiple 2-number factors is mine when the RHS has more then 2 prime factors---the aforementioned video had exactly 2 prime factors--, and we solve the problem for each of these **valid** factors [those w/ unequal numbers] to see which yields integer solutions)
Since a+b > a - b, the only valid factor is 8*2 (in general, there may be more than one valid 2-number factor and we need to try each for possibly multiple integer solutions for x).
Thus (a+b)(a-b) = 8*2 ==> a+b = 8, a-b = 2 ==> 2a = 10 ==> a = 5 ==> 5^{x/2} = 5
==> x/2 = 1 ==> x = 2 as the only integer solution (a verification---a given--by plugging in x=2 in the original equation, proves it).
nice
Brilliant this is another way of looking at math problems like another idea it could be use I some problems so as many ideas like this we have somewhere in our head we would be better at solving problems in general
3^2 + 4^2 = 5^2
직깍상각형 길이자승의 법칙을 알면 이 문제에 한정지으면 금방 풀리는 문제이지만
수학적으로 푸는과정은 어렵네요..
It’s also a really nice equation because it’s a different form of the Pythagorean triple 3, 4, 5
I didn't understand a word of this video but I got the right answer in about 3 seconds. This has been my problem with maths all my life.
its really tricky but maths was fun.😜
Fantastic analysis of the problem!!!
Thank you 🧡
Fast analytic method after guessed x=2:
Let f(x) = 5^x - 3^x
for x < 0, we don't need to worry too much because 5^x < 3^x < 1, which means f(x) must be negative.
f(0) = 0
f'(x) = 5^x * ln(5) - 3^x * ln(3) which is always > 0 when x >= 0
So f(x) is strictly increasing when x >= 0
x=2 is the only solution.
Other than by Inspection - in a few seconds - or trial and error - in a few seconds, you could write this as:
5^x - 3^x = 16 = 4^2 (or option 2 as 2^4). Let's try the first option as x=2 then the seconds as x =4.
So rearrange and you get 5^x = 3^x +4^2 (implying x = 2) or 5^x = 3^x +2^4 (implying x = 4),
now 5^2 = 3^2 + 4^2 (25 = 9 + 16) but 5^4 3^4 + 2^4 (as 625 81 + 16).
So if you just guessed x= 2 - or you knew a 3,4, 5, right angle triangle - the answer would fall out immediately. You could also attempt to solve this for 16 = 2^4, but if you try for x = 4 you would see it doesn't work as 5^4 3^4 + 2^4 so x must equal 2 is the only answer.
This would be more interesting if you can solve, say, 5ˣ − 3ˣ = 13.
How?
That was really interesting bro , I appreciate that
Thank you!
Man, I blew the answer big time. Thank you for walking me through it. 👍👍👍
Glad to help!
An incredible way of solving
Thank you! 🤩
It's clear that they diverge so there's just one solution to be found. Some quick trial-and-error gets you x=2.
my math teacher once told me.
"If we were supposed to trial and error, then we wouldn't have math"
x=2 est évident
Ensuite c’est évident que 5^x augmente plus vite que 3^x, donc la différence sera toujours plus grande que 16
I don't think this is really the correct way to solve this equation - it's just one way of brute force - not purely mathematical. I think it is better to consider that 5 to the same power (except zero) always has 5 as the last digit - because xx5 * 5 = xxx5 etc. On the other hand 3^x must have the last digit 9 to give the number with last digit 6 on the right! And that's it! Now we only have to prove when 3 ^ x gives the number with the last digit 9? And this only happens with the pattern: 3^(2 + 4 * n). And because 5^(2 + 4 * n) - 3^(2 + 4 * n) is an increasing function > 16 for n > = 1 than n must be 0. So this is the correct solution => 5^(2+0) - 3^(2+0) = 16 => x = 2!
This solution is in no way more algebraic or correct than guessing 😂
X = 2: 5 squared = 25 and 3 squared = 9, 25 - 9 = 16.
Too easy. Thank you for your efforts. May you and yours stay well and prosper.
Thanks for that! ❤️
Although the process did not seem rigid but it is well explained. I am quite interested to know that such a "difficult" problem can be solved by trial and error !
Glad to hear that
Are there any complex solutions (other than 2+0i)? Thanks.
Yes, there are, x=2+2kπi is a solution for each integer k.
Setting y=e^x, the equation gets y^(ln 5)-y^(ln 3)=16.
It is easy to see that y will be the same if x changes over 2πi.
We can go a bit further. Let x=a+bi. Then 5^x and 3^x are somewhere on the circles with radii 5^a and 3^a, respectively. The phase difference depends on the imaginary part. We could try to solve the system
5^a cos(b ln(5)) - 3^a (cos b ln(3)) = 16
5^a sin(b ln(5)) - 3^a (sin b ln(3)) = 0
... but tbh I don't really expect to find any x with real part other than 2.
I saw and got x=2 instantly.Thank U.
YES ! x =2. Answer given within 10-15 seconds .
答え方 16=4^2 そう直感でx=2を代入
5^2ー3^2=16 正解です ここの説明はムズ
ただ 0
I just noticed the equation is: 5^x = 3^x + 4^2, and the common 3x4x5 triangle works to solve this via the Pythagoras theorem when x=2. Also, it appears to only have 1 answer, because 5^x is increasing and 3^x is decreasing (but less than 5^x) and 4^2 is a constant. Hence, the only answer is x=2. No other maths required. boom. answer in like 5sec.
The triangle way to solve is amazing. I have another.
5^x - 3^x = 16
25^(x/2) - 9^(x/2) = 16
if 25-9=16
x/2 has to be equal to 1
x/2=1 then x=2
Fist hint 5^x-3^x=16=4^2. Second clue 3,4 and 5 square are related by 3^2+4^2=5^2. In this case x=2 gives first solution.
My preferred method is to change the equation to ((5^(x/2))^2)-((3^(x/2))^2)=8×2, then substitute, respectively, u and v for 5^(x/2) and 3^(x/2). We then have (u^2)-(v^2)=8×2. This factors to (u+v)(u-v)=8×2, which makes for a relatively easy equation system of u+v=8 and u-v=2. U and v are, respectively, 5 and 3, which are easy to solve for each exponent of x/2.
I don't like guess and check. I like definite general rules for a given equation form of any type.
nice
Thank you professor
You are very welcome
Ça vient direct pour ceux qui maîtrisent le théorème de Pythagore concernant le triangle rectangle avec angle droit a²+b² = c²
A partir de là la solution se voit
16 étant le carré de 4 on ne fait qu'appliquer le théorème de Pythagore
" Find a solution "
Me : 😃
" Find the number of real solutions "
Me : 😵💫😵
Good job on finding the solution! This whole video made me incredibly anxious! 🤯
😁😜
let assume original equation is 5^x-3^x=a chose a number for x then put in right side of this equation that i got from original equation x=ln(a-3^x)/ln5 and get new X and repeat those steps, after 3 or 4 steps you get approximate answer. in this way it doesn't mater a is equal 16 or any number
Very easy after seeing the question. I get answers x=2
It would be nice to see this used to solve for any given x. Solving this for x in terms of y would be very cool. You can eyeball this and see that it's 2, but isolating x and solving for it for any given y would be nice.
Cannot be done
Thx, good description.
Glad it was helpful!
A very easy question everyone can do. You are elaborating it a way more than the distance between The Sun and the Pluto.
😱🤪😂
You made math so hard. Anyone just looking at the equation can figure it out without doing all that work
Right!
I got it. Very clear proof but no mention of continuity at any time...
5^x - 3^x = 4^2. (3,4,5) are known pythagorian numbers, ergo x = 2. Done. This is beyond easy. Also it is known that the exponential function is monotonous, meaning there won't be any other value besides x = 2 that fulfills the same role as x = 2
considering 3=9^(½) and 5=25^(½)
so we can write the equation like this:
25^(x÷2)-9^(x÷2)=16
but we know that 25¹-9¹=16
so x÷2=1
consequently x=2
ここの説明じゃムズで厄介
xを0、1、2と代入すればよい
16=4^2と考え2を直感で代入もOK
Solution
Doing x = a, we have
5ª - 3ª = 16
5ª - 3ª = 25 - 9
5ª - 3ª = 5² - 3²
5ª - 5² = 3ª - 3²
5ª - 5² = 0 => 5ª = 5² => a = 2
or
3ª - 3² = 0 => 3ª = 3² => a = 2
Being a = 2, then x = 2.
It is in the form of A^2 - B^2
Then in the place of ^2 x is there then x= 2 ..
(A+B)(A-B)=16
(5+3)(5-3)=16
16=16.
you have a soothing and friendly voice :D
Thank you! 🥰
This is a simple equation, just write 5=3+2 and then apply binomial Newton, then remove 3^x. Simplify by 2 from both sides, then the remaining sum should be 8. The only case is ×=2. Since the formula increases with x.
When ever I saw that question I substituted x=2 I got the answer and fast forwarded the video to verify my answer. I got correct answer that is x=2 and I got my answer in less than 5 seconds.I know many of got the answer like me.
😂😂😂😂 bro you made this problem complicated 😂😂😂😂 idk why cuz you can use this method instead
Here it is :-
First write 5^x -3^x = 16
Now write 16 as 5²-3²
After comparing the base we get x =2 easyyyyyy
AMAZING THANK SIR
Fascinating! I wonder if you could use this and some nice trick with pythagorean triples to somehow prove FLT?
I solved it in my head in less than 30 seconds.
Just reminded me of the 3 4 5 triangle, so I just got it right away.
a naive way in Python
for i in range (0,100):
if 5**i - 3**i == 16:
break
print(i)
output: 2
👍I guessed the solution at the starting of video. But finding how much possible solutions this equation could have was important part, so that we can guess and find all possible solutions. Good tutorial, keep it up.👍