The Power of X

Starting with $9^x=6^x+4^x$ , we divide both sides by $4^x$ .

$\begin{aligned}\begin{aligned}\left(\frac94\right)^x&=&\left(\frac32\right)^x+1\\\left(\frac32\right)^{2x}-\left(\frac32\right)^x-1&=&0\end{aligned}\end{aligned}$

Let $y=\left(\dfrac32\right)^x$ . This makes the above equation become $y^2-y-1= 0 \longrightarrow y = \dfrac{1\pm\sqrt{5}}{2}$ .

Since the exponential functions $\left(\dfrac32\right)^{2x}$ and $\left(\dfrac32\right)^x$ are defined over positive numbers, the value of $\left(\dfrac32\right)^{2x}$ and $\left(\dfrac32\right)^x$ are positive. Thus, $\dfrac{1-\sqrt5}2$ is cannot be the value of $\left(\dfrac32\right)^x$ implying that the value of $\left(\dfrac32\right)^x=\dfrac{1+\sqrt5}2$ satisfying is $x=\boxed{\log_{\frac32}{\left(\dfrac{1+\sqrt5}2\right)}}$

We begin with $9^x = 6^x + 4^x$ .

First, express each integer as it's prime factorization: $(3 \cdot 3)^x = (2 \cdot 3)^x + (2 \cdot 2)^x$

Then use the fact that $(ab)^x = a^xb^x$ to rewrite the equation: $3^x \cdot 3^x = 2^x \cdot 3^x + 2^x \cdot 2^x$

Factor out $2^x$ from the RHS: $3^x \cdot 3^x = 2^x (3^x + 2^x)$

Divide both sides by $2^x$ and simplify: $(\frac{3}{2})^x \cdot 3^x = 3^x + 2^x$

Now guess that the solution is of the form $log_{\frac{3}{2}} a$ to get: $a \cdot 3^{log_{\frac{3}{2}} a} = 3^{log_{\frac{3}{2}} a} + 2^{log_{\frac{3}{2}} a}$

Subtract both sides by $3^{log_{\frac{3}{2}} a}$ to get: $(a -1) \cdot 3^{log_{\frac{3}{2}} a} = 2^{log_{\frac{3}{2}} a}$

Divide both sides by $2^{log_{\frac{3}{2}} a}$ and simplify: $(a -1) \cdot (\frac{3}{2})^{log_{\frac{3}{2}} a} = 1$

Simplify using properties of logs and simple algebra to get: $a^2 - a - 1 = 0$

By the quadratic equation: $a = \frac{1 \pm \sqrt{5}}{2}$ , but since all the integers are positive, $a$ must be postive.

Finally, arrive at the answer: $x = log_{\frac{3}{2}} a = log_{\frac{3}{2}}(\frac{1 \pm \sqrt{5}}{2})$