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 besomewhere 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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