Problem 1

Solve for all real values of x such that

9^x + 12^x = 16^x

Problem 2

Solve for all real values of x such that

(8^x – 2^x)/(6^x – 2^x) = 2

As usual, watch the video for a solution.

Two Exponential Equations

Or keep reading.
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“All will be well if you use your mind for your decisions, and mind only your decisions.” Since 2007, I have devoted my life to sharing the joy of game theory and mathematics. MindYourDecisions now has over 1,000 free articles with no ads thanks to community support! Help out and get early access to posts with a pledge on Patreon.

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Answer To Exponential Equation 8 to x minus 2 to x

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

First divide each term by 9^x

9^x + 12^x = 16^x
9^x/9^x + 12^x/9^x = 16^x/9^x
1 + (12/9)^x = (16/9)^x
1 + (4/3)^x = ((4/3)²)^x
1 + (4/3)^x = (4/3)^2x

Then let u = (4/3)^x so the equation reduces to a quadratic equation.

1 + u = u²
0 = u² – u – 1

Using the quadratic formula we get two roots

u = (1 ± √(1 – 4(1)(-1)))/2

u = (1 + √5))/2
u = (1 – √5))/2

Since u = (3/2)^x, this is greater than 0 for real values of x. So we reject the second root which is negative. Thus:

u = (1 + √5))/2

So u is equal to the golden ratio! Now we have one step left: to solve for x. We can do that with the help of the natural logarithm.

u = (4/3)^x
ln u = ln[(4/3)^x]
ln u = x ln(4/3)
x = ln u/(ln (4/3))
x = [ln((1 + √5)/2)]/(ln (4/3))
x ≈ 1.673

The equation is challenging at first. But with a few clever tricks it is possible to find a golden answer!

Problem 2

This equation is much harder and we unfortunately cannot use the same trick as in problem 1. So let’s simplify it carefully and investigate its behavior.

Clearly x ≠ 0 because the fraction would be undefined:

(8⁰ – 2⁰)/(6⁰ – 2⁰)
= (1 – 1)/(1 – 1)
= 0/0

Also x = 1 is a solution

(8¹ – 2¹)/(6¹ – 2¹)
= (8 – 2)/(6 – 3)
= 4/2
= 2

Are there any other real solutions? The answer is no!

If x ≠ 0, then we have:

(8^x – 2^x)/(6^x – 2^x)
= (2^3x – 2^x)/(2^x3^x – 2^x)
= 2^x(2^2x – 1)/[3^x(2^x – 1)]
= 2^x(2^x – 1)(2^x + 1)/[3^x(2^x – 1)]

Since x is not 0, we can cancel 2^{x – 1} in the numerator and denominator.

2^x(2^x + 1)/3^x
= (2^2x + 2^x)/3^x
= (4^x + 2^x)/3^x
= (4/3)^x + (2/3)^x

Define a function for all real values of x:

f(x) = (4/3)^x + (2/3)^x – 2

Every solution of the original fraction will be a root of this equation. What are the roots of this equation? Calculate the second derivative:

f”(x) = (4/3)^x(ln (4/3))² + (2/3)^x(ln (2/3))²

Every term is strictly greater than 0, so the entire 2nd derivative is strictly greater than 0. This means the original function is convex and has at most 2 distinct roots.

Clearly x = 1 is a root. But we also have x = 0 is a root:

(4/3)⁰ + (2/3)^x – 2
= 1 + 1 – 2
= 0

Every root of the original fraction equation would be a root of this function. But this function has two roots x = 1 and x = 0. Since x ≠ 0 in the original fraction, the only real root is x = 1.

References

Similar to problem 1 on Socratic
https://socratic.org/questions/how-do-you-solve-6-x-4-x-9-x

Problem 2 @MathBooster video integer solutions
https://www.youtube.com/watch?v=X3OiHW98QTE

Problem 2 Math StackExchange
https://math.stackexchange.com/questions/4699669/frac8x-2x6x-3x-2-solve-for-x

Problem 2 strictly convex
https://math.stackexchange.com/questions/2112108/prove-a-convex-and-concave-function-can-have-at-most-2-solutions

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