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

Infinitesimal's picture

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, you get
Simplifying the fractions you get
That can be written as
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
Multiplying both sides times k,
Solving that quadratic equation gives you two real solutions for k.

Infinitesimal's picture

There is something misleading hahaha.

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

Jhun Vert's picture

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

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

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

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

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

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


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