The function $f(x)=4^x+6^x-9^x$ is such that $f(0)=1&gt;0, f(1)=1&gt;0, f(2)=-29$ and next $g(x)=(4/9)^x+(6/9)^x-1 \implies f'(x)&lt;0$ for all real values of $x$. So $g(x)$ being monotonic the equation $$4^x+6^x=9^x$$ has exactly one real solution. The question is whether this real root can be found analytically by hand.
So here is the problem      Solve $$4^x + 6^x = 9^x$$ for $x$.   I am trying to find its real solution.  I was trying in this way!! $$6^x\left( \frac {4^x}{6^x} + 1\right) = 9^x$$ $$\left({\frac 23}\right)^x +1=\left({\frac 32}\right)^x$$ but I'm stuck . . . Will anyone help ?
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