# Exponential Function: $4^x + 6^x = 9^x$ Let $x$ be a real number that satisfy the equation

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

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

### Dividing both sides by 6x,

Dividing both sides by 6x, you get
4x/6x+1=9x/6x
Simplifying the fractions you get
2x/3x+1=3x/2x
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,
k2+k=1
Solving that quadratic equation gives you two real solutions for k. 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.
$4^x + 6^x = 9^x$

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

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

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

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

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

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

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

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

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

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

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

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

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

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