Question

Solve the equation `(log_2x)/(log_2 2x)=(log_8 4x)/(log_4 8x)`?

Answer 1

`x=2` or `1/16`

Explanation:

As `log_xy=log_ay/log_ax`

`log_8 4x=(log_2 4x)/(log_2 8)=(log_2 4x)/(log_2 2^3)=1/3log_2 4x`

and `log_4 8x=(log_2 8x)/(log_2 4)=(log_2 8x)/(log_2 2^2)=1/2log_2 8x`

Hence `(log_2x)/(log_2 2x)=(2log_2 4x)/(3log_2 8x)=(2(log_2 4+log_2x))/(3(log_2 8+log_2x)`

or `(log_2x)/(1+log_2x)=(2(2+log_2x))/(3(3+log_2x))`

Let `log_2x=a`, then this becomes

`a/(1+a)=(2(2+a))/(3(3+a))` or `3a(3+a)=2(2+a)(1+a)`

or `3a^2+9a=2a^2+6a+4`

or `a^2+3a-4=0` i.e. `(a+4)(a-1)=0`

Hence `a=-4` or `1`

if `a=-4`, `log_2x=-4` i.e. `x=2^(-4)=1/16`

and if `a=1`, `log_2x=1` i.e. `x=2^1=2`