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$4^{x} + 6^{x} = 9^{x}$ $4^x+6^x=9^x$

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Answer 1 · 2018-06-25T07:11:25

$4^{x} + 6^{x} = 9^{x}$ $4^x+6^x=9^x$

$\Rightarrow \frac{4^{x}}{6^{x}} + 1 = \frac{9^{x}}{6^{x}}$ $=>4^x/6^x+1=9^x/6^x$

$\Rightarrow {(\frac{4}{6})}^{x} + 1 = {(\frac{9}{6})}^{x}$ $=>(4/6)^x+1=(9/6)^x$

$\Rightarrow {(\frac{2}{3})}^{x} + 1 = {(\frac{3}{2})}^{x}$ $=>(2/3)^x+1=(3/2)^x$

Let ${(\frac{3}{2})}^{x} = y$ $(3/2)^x=y$

So $x log (\frac{3}{2}) = log y$ $xlog(3/2)=logy$

$\Rightarrow x = \frac{log y}{log 1.5}$ $=>x=(logy)/log1.5$

Again

$y - \frac{1}{y} = 1$ $y-1/y=1$

$\Rightarrow y^{2} - y - 1 = 0$ $=>y^2-y-1=0$

$\Rightarrow y = \frac{1 \pm \sqrt{1 - 4 \cdot 1 (- 1)}}{2}$ $=>y=(1pmsqrt(1-4*1(-1)))/2$

$\Rightarrow y = \frac{1 \pm \sqrt{5}}{2}$ $=>y=(1pmsqrt5)/2$

So to get the value of $x$ $x$ we can use the positive root of $y$ $y$ as $x$ $x$ is related with $log y$ $logy$ .

Hence $x = \frac{log (\frac{\sqrt{5} + 1}{2})}{log 1.5}$ $x=log((sqrt5+1)/2)/log1.5$