A Difficult Exponential Question

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[lanewaygarden]

Excellent explanation and well set out exact solution

[crtwrght]

I saw this question in the thumbnail and I loved it. The fact that just two lines of working in it suddenly transforms into an elementary quadratic is delightful.

[ricardofraser]

anyone else noticed that T is the golden ratio because R(n+2)=R(n+1)+R(n) which is a recurrence relation for the Fibonacci sequence and T^2-T-1=0 is the characteristic equation.

[mrIceblink]

The real-valued solution is approx 2.15651

The other solution, which is complex-valued, is also valid -2.15651 + 14.0788i (where you take the logarithm of (1 - sqrt(5))/2)

[psy]

Thank you for this thought-provoking question. Once the division of 16^x was carried out, the solution was rather straightforward. Well done.

[RexxSchneider]

Alternatively, you could divide through by 25^x, giving (16/25)^x + (20/25)^x = 1. Then let r = (4/5)^x. That gives r^2 + r = 1 or r^2 + r - 1 = 0. That has solutions r = (√5 - 1)/2 and (-√5 - 1)/2. Since taking logs of negative numbers leads to complex solutions, we can take r = (√5 - 1)/2 to give a real solution when we take logs of (4/5)^x = (√5 - 1)/2.
This solution is x = ln((√5 - 1)/2) / ln(4/5) ≈ 2.1565, which corresponds to the solution found in the video. That is the case because the numerators for the two paths to the solution are 1/ɸ and ɸ, respectively, and the denominators are also reciprocals of each other. Note that log(1/x) / log(1/y) = log(x) / log(y).

[cufflink]

Quick note: It's not necessary to use natural logarithms (ln's) here, as the explanation seems to imply. Logs to any base at all, including base 10, work just as well, since you're simply dividing one logarithm by another in the same base.

[chooshchoosh]

Impressed that he resisted the distraction of mentioning the Golden Ratio

[bosorot]

a little short cut here is let p=5^x and q=4^x . you can re-write it to 0 = p^2-pq-q^2 . Then just plug in to quadratic formula . a=1, b= -q, c=-q^2. the answer is p=q . golden ratio.

[RSVikingJohn]

Maybe I was luckly and found it easy, but here was my thought process.

Figuring things out:
The first immediate thought I had was to divide it into groups partly because we have two pretty square numbers and because maybe it would show me some nice connections:
16^x + 20^x = 25^x
(4^x)(4^x) + (4^x)(5^x) = (5^x)(5^x)
Dividing by 5^x because I notice I can factorize the expression in regards to 4^x on the leftside afterwards,
(4^x)((4/5)^x) + (4^x) = (5^x)
(4^x)( ((4/5)^x) + 1) = (5^x)
Wow okay, now I notice that the equation can be easily substituted if I divide by 5^x again
(4/5)^x ( ((4/5)^x) + 1) = 1

Substituting:
Let v=(4/5)^x. The equation then becomes,
v ( v + 1) = 1

Solving the substituted equation, v:
Expanding and rearranging,
v^2 + v - 1 =0
Using the quadratic formula to solve for possible values of v:
v_- = 1/2 (-1 - \sqrt(5));
v_+ = 1/2 (-1 + \sqrt(5))
I remember when I was in highschool I forgot to check if the solution satisfied the original substitution, so I better check now. Looking at it, I see that v cannot be negative since v=(4/5)^x always greater than 0, and v_- is a negative number. Thus I have seen that v_- is not a legitimate solution to the original problem.
v = 1/2 (-1 + \sqrt(5)) = (4/5)^x

Solving for the original problem, x:
v = 1/2 (-1 + \sqrt(5)) = (4/5)^x
Taking the logarithm:
-ln(2) + ln(-1+\sqrt(5)) = x (ln(4) - ln(5))
Solving the equation above for x we get,
x = ( -ln(2) + ln(-1+\sqrt(5)) )/( ln(4) - ln(5) )

[proberts]

"39 easy steps to get an answer that doesn't make sense". And this is why I never made it past Algebra. I'm glad that there are people in the world who have an affinity for mathematics.

[bjbell]

Darn, I was working it correctly, but I missed one step. Back in 1991 I could have easily done it (that's when I earned my B.S. degree in mathematics) but 30 years later, I forgot so much.

[frentz]

Divide on both sides by 25^x, and set T = (4/5)^x. The new equation is T^2 + T = 1, or T^2 + T - 1 = 0. Now solve the quadratic, and then solve for x.

[FahlmanCascade]

I saw the video preview on the YouTube front page and made a point to solve the question before coming to watch this video. I got x = log(ɸ) / log(5/4). I had to try a few things before I discovered a working set of algebraic manipulations. What I came up with was not exactly short. I'm curious to see whether there's a more direct path to the answer than the one I found.

Well, I'm back, and the solution shown was almost exactly the one I found, no shorter.

The explanation as to why we should exclude the negative solution to the quadratic part of the problem could have been clearer. We know that we are raising the positive value, 5/4, to some power. We cannot get a negative answer when we exponentiate a positive base, so we need only to consider the positive answer.

[jpl]

Nicely done, but I'd better do this : dividing by 20^x, I obtain (5/4)^x - (4/5)^x - 1 = 0... then replacing (5/4)^x by a, it gives a - 1/a - 1 = 0, or a^2 - a - 1 = 0.
This Fibonaci equation gives the positive anxwer ɸ = (1+√5)/2, leading to the final answer x = ln (ɸ) / ln (5/4). Thanks to the authors !

[YodaWhat]

OR, you can *solve the given equation experimentally* a digit at a time, by numerical methods.

[M-F-H]

Much simpler, from 16^x + 20^x + 25^x => 1 + (20/16)^x = (25/16)^x = (5/4)^2x =: phi², so: 1 + phi = phi² which defines the Golden Ratio phi(+-) = (sqrt(5)+-1)/2, thus (5/4)^x = phi, or x = log phi / log(5/4) -- two solutions which differ just in sign because phi(-)= 1/phi(+). Takes less than 30 seconds....

[attilakiss]

a=4^x, b=5^x. Solve equation b*b-ab-a*a=0 (second order). Result b=((1-sqrt(5))/2)*a, so you have b/a. Quite straightforward.

[pythagorasaurusrex]

Just my 50 cents: You can find similar equations here on youtube. All are solved the same way: Divide by the highest power, to get rid of one x, substitude the power in such a way you get a polynomial. Solve the polynomial, pick the positive solution and re-substitude by taking the logs accordingly... Voila!
Anyway... once you got rid of the highest power by dividing with it, the rest is quite straight forward.

EDIT: you can find 3^x+9^x=27^x or 4^x+6^x=9^x here on YT. All solved the same way.

[ffggddss]

The key to solvability of this, turns out to be that 16, 20, 25 are in geometric sequence.
Any three (different!) positive numbers that are in geometric sequence, would, in place of 16, 20, 25, lead to a solution by this method.

Slightly simpler, after dividing both sides by 16ˣ, group the fractions like this:
1 + (20/16)ˣ = (25/16)ˣ
then do the reduction on the LHS and re-express the RHS:
1 + (5/4)ˣ = (5²/4²)ˣ = (5/4)²ˣ = [(5/4)ˣ]²

and it's on to the quadratic in (5/4)ˣ, leading to the solution.

Fred