Solve the following ?

Question

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

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Answer 1 · 2018-06-25T07:11:25

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

$=>4^x/6^x+1=9^x/6^x$

$=>(4/6)^x+1=(9/6)^x$

$=>(2/3)^x+1=(3/2)^x$

Let $(3/2)^x=y$

So $xlog(3/2)=logy$

$=>x=(logy)/log1.5$

Again

$y-1/y=1$

$=>y^2-y-1=0$

$=>y=(1pmsqrt(1-4*1(-1)))/2$

$=>y=(1pmsqrt5)/2$

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

Hence $x=log((sqrt5+1)/2)/log1.5$