How to solve indicial equations with different bases

Solve the indicial equation

$4^\frac{1}{x}-6^\frac{1}{x}=9^\frac{1}{x}$

To solve the equation, simply divide through by any of the term since the indices are all given in different bases. So we divide by $9^\frac{1}{x}$.

$\frac{4^\frac{1}{x}-6^\frac{1}{x}}{9^\frac{1}{x}}=1$

$(\frac{4}{9})^\frac{1}{x}-(\frac{6}{9})^\frac{1}{x}=1$

Express the bases in homogenous form.

$(\frac{2^2}{3^2})^\frac{1}{x}-(\frac{2}{3})^\frac{1}{x}=1$

$(\frac{2}{3})^\frac{2}{x}-(\frac{2}{3})^\frac{1}{x}=1$

Solve the equation using substitution method. Let $p=(\frac{2}{3})^\frac{1}{x}$................................(*)

The equation is now a quadratic equation

$p^2-p=1$

Solve the equation using quadratic formula method and you will obtain two roots of the form

$p_{1,2}=\frac{1\pm\sqrt{5}}{2}$

Substitute back the value of $p_1=\frac{1+\sqrt{5}}{2}$ into equation (*) and find $x$. 

$(\frac{2}{3})^\frac{1}{x}=\frac{1+\sqrt{5}}{2}$

Take the $\log$ of both sides so that we can make $x$ subject of formula. 

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

$x=\frac{\log\frac{2}{3}}{\log(\frac{1+\sqrt{5}}{2})}=\frac{\log2-\log3}{\log(1+\sqrt{5})-\log2}$. 

$\approx{-0.842591738...}$

There is no real value for $p_2$ because $p_2=\frac{1-\sqrt{5}}{2}<0$