# Misc Solution: Solve exponential equation $$6^x + 4^x = 9^x$$

Given the equation $$6^x + 4^x = 9^x$$, solve for all $$x$$.

## Solution

The idea is to divide the whole equation by $$4^x$$:

$\begin{array}{rrl} & 4^x + 6^x &= 9^x\\ \Leftrightarrow & 1 + \frac{6^x}{4^x} &= \frac{9^x}{4^x}\\ \Leftrightarrow& 1 + \left(\frac{3}{2}\right)^x &= \left(\left(\frac{3}{2}\right)^2\right)^x = \left(\left(\frac{3}{2}\right)^x\right)^2\\ \end{array}$

In this way we get a quadratic formula in $$q=\left(\frac{3}{2}\right)^x$$:

$\begin{array}{rrl} & 1 + q &= q^2\\ \Leftrightarrow &q^2 - q - 1 &= 0\\ \Leftrightarrow &(q-\frac{1}{2})^2 &= \frac{5}{4}\\ \Leftrightarrow &q &= \frac{1}{2}\pm\frac{\sqrt{5}}{2}\\ \Leftrightarrow &q &= \frac{1\pm \sqrt{5}}{2}\\ \end{array}$

Since $$q=\left(\frac{3}{2}\right)^x>0$$, we can ignore the negative solution and proceed with the positive (golden ratio) $$q = \frac{1+ \sqrt{5}}{2}$$. Setting both variants for $$q$$ equal yields

$\begin{array}{rrl} &\left(\frac{3}{2}\right)^x &= \frac{1+ \sqrt{5}}{2}\\ \Leftrightarrow &x\cdot \ln\left(\frac{3}{2}\right) &= \ln\frac{1+ \sqrt{5}}{2}\\ \Leftrightarrow &x &= \frac{\ln\frac{1+ \sqrt{5}}{2}}{\ln\left(\frac{3}{2}\right)} \\ &&= 1.1868143\dots \end{array}$

## Solution using Newton Raphson method

We can solve this problem also numerically instead, using Newton Raphson method.

Let $$f(x) = 4^x + 6^x - 9^x$$, then $$f'(x) = 4^x \ln 4 + 6^x \ln 6 - 9^x \ln 9$$. Since we want the solution $$x$$, for which $$f(x) = 0$$, we try some small values: $$f(1) = 1$$, $$f(2) = -29$$. That means $$f(x)=0$$ must be somewhere between 1 and 2, so we start with $$x_0=1.5$$:

The next better estimate for $$x$$ is given by $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$:

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