Question

Solve the following ?

`4^x+6^x=9^x`

Answer 1

`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`