# Exponential Function: $4^x + 6^x = 9^x$

Submitted by Infinitesimal on August 30, 2016 - 11:29am

Let $x$ be a real number that satisfy the equation

$4^x + 6^x = 9^x$

What is the value of $\left(\dfrac{2}{3}\right)^x$ ?

- Add new comment
- 671 reads

## Dividing both sides by 6x,

Dividing both sides by 6

^{x}, you get4

^{x}/6^{x}+1=9^{x}/6^{x}^{}Simplifying the fractions you get

2

^{x}/3^{x}+1=3^{x}/2^{x}^{}That can be written as

(2/3)

^{x}+1=(3/2)^{x}Now, if the number you want is (2/3)

^{x}=k, then (3/2)^{x}=1/k ,which means the equation can be written as

k+1=1/k

Multiplying both sides times k,

k

^{2}+k=1Solving that quadratic equation gives you two real solutions for k.

## There is something misleading

There is something misleading hahaha.

Do the two real solutions for k BOTH give REAL solutions for $x$ ?

## I am not used anymore to

I am not used anymore to calculate this type of equation, the result of relying too much in

`SHIFT + SOLVE`

of Casio. I did not say doing`Shift + Solve`

is bad, it is actually highly recommended in my line of work, hehehe. Anyway, allow me to solve this, and I am actually surprised that solving for x is more complex than solving for (2/3)^{x}. Here is my take based on the suggestion of KMST.$4^x + 6^x = 9^x$

$\dfrac{4^x}{6^x} + \dfrac{6^x}{6^x} = \dfrac{9^x}{6^x}$

$\left( \dfrac{4}{6} \right)^x + 1 = \left( \dfrac{9}{6} \right)^x$

$\left( \dfrac{2}{3} \right)^x + 1 = \left( \dfrac{3}{2} \right)^x$

$\left( \dfrac{2}{3} \right)^x + 1 = \dfrac{1}{\left( \dfrac{2}{3} \right)^x}$

$\left( \dfrac{2}{3} \right)^{2x} + \left( \dfrac{2}{3} \right)^x = 1$

$\left( \dfrac{2}{3} \right)^{2x} + \left( \dfrac{2}{3} \right)^x - 1 = 0$

By Quadratic Equation

$\left( \dfrac{2}{3} \right)^x = \dfrac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)}$

$\left( \dfrac{2}{3} \right)^x = \dfrac{-1 \pm \sqrt{5}}{2}$

$\left( \dfrac{2}{3} \right)^x = \dfrac{-1 + \sqrt{5}}{2}$

$\log \left( \dfrac{2}{3} \right)^x = \log \dfrac{-1 + \sqrt{5}}{2}$

$x \log \left( \dfrac{2}{3} \right) = \log (\sqrt{5} - 1) - \log 2$

$x (\log 2 - \log 3) = \log (\sqrt{5} - 1) - \log 2$

$x = \dfrac{\log (\sqrt{5} - 1) - \log 2}{\log 2 - \log 3}$ ← a real number

For

$\left( \dfrac{2}{3} \right)^x = \dfrac{-1 - \sqrt{5}}{2}$

$\log \left( \dfrac{2}{3} \right)^x = \log \dfrac{-1 - \sqrt{5}}{2}$

$x \log \left( \dfrac{2}{3} \right) = \log (-\sqrt{5} - 1) - \log 2$

$x (\log 2 - \log 3) = \log (-\sqrt{5} - 1) - \log 2$

$x = \dfrac{\log (-\sqrt{5} - 1) - \log 2}{\log 2 - \log 3}$ ← underfined

Hence,

$\left( \dfrac{2}{3} \right)^x = \dfrac{\sqrt{5} - 1}{2}$ ← this is my answer.

Please be gentle with me, hehehe. Although I hope I was able to consider everyhting, maybe I miss simething. As I've said, I am no longer used to this type of approach. A decimal number from calculator is more than acceptable for me.

## Add new comment