Question

How do you solve `log_2((3x-2)/5)=log_2(8/x)`?

Answer 1

`x=4`

Explanation:

`log_"2"((3x-2)/5)=log_"2"(8/x)`

`D_"f":x in RR_"+"^"*"`

`(3x-2)/5=8/x`

`3x²-2x=40`

`3x²-2x-40=0`

`Δ=4+4*3*40`

`Δ=484`

`x_"1"=(2-sqrt(484))/6`

`x_"2"=(2+sqrt(484))/6`

`cancel(x_"1"=-10/3)` impossible due to our `D_"f"`

`x_"2"=4`

\0/ here's our answer!

Answer 2

`color(maroon)(x = 4`

Explanation:

`log_2 ((3x-2)/5) = log_2 (8/x)`

`(3x-2)/5 = (8/x), " Removing Log on both sides"`

`3x^2 - 2x = 40`

`3x^2 - 12x + 10x - 40 = 0`

`3x (x - 4) + 10 (x - 4) = 0`

`(x - 4) * (3x + 10) = 0`

`x = 4, color(crimson)(cancel(-10/3)`

`color(maroon)(x = 4`