Question

How do you solve `(log_2(2x-3) / log_2(x)) - log_x(x+6) + (1/log_ (x+2)(x)) = 1`?

Answer 1

`x^3+8x^2+10x+3=0`

Explanation:

`(log_2(2x-3)/log_2 x)-log_x(x+6)+(1/log_(x+2) x)=1`
`log_x(x+6)=(log_2 (x+6))/(log_2x)`
`log_(x+2) x=(log_2 x)/(log_2(x+2))`
`(log_2(2x-3)/log_2 x)-log_2(x+6)/(log_2 x)+log_2(x+2)/log_2 x=1`
`(log_2(2x-3)-log_2(x+6)+log_2(x+2))/(log_2 x)=1`
`log_2(2x-3)-color(red)(log_2(x+6)+log_2(x+2))=log_2 x`
`color(green)(log_2(2x-3)-log_2 (x+6)*(x+2))=log_2 x`
`log_2 ((2x-3)/((x+6)(x+2)))=log_2 x`
`(2x-3)/((x+6)(x+2))=x`
`x(x^2+2x+6x+12)=2x-3`
`x^3+2x^2+6x^2+12x-2x+3=0`
`x^3+8x^2+10x+3=0`