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Solve for x, $4^x+6^x=9^x$

We have, ${({2^x})^2} + {2^x}{.3^x} = {({3^x})^2}$

Dividing by $2^x.3^x$,

${\left( {\frac{2}{3}} \right)^x} + 1 = {\left( {\frac{3}{2}} \right)^x}$

Let, ${\left( {\frac{3}{2}} \right)^x} = t$

We have, $\frac{1}{t} + 1 = t$

Or, ${t^2} - t - 1 = 0$

So, $t = \frac{{1 \pm \sqrt {1 + 4} }}{2} = \frac{{1 \pm \sqrt 5 }}{2}$

Since, $t = {\left( {\frac{3}{2}} \right)^x} > 0,t = \frac{{1 + \sqrt 5 }}{2}$

We have, ${\left( {\frac{3}{2}} \right)^x} = \frac{{1 + \sqrt 5 }}{2}$

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

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

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