Please help! I can't figure this out!

Question

log₂x-2=√log₂x-2

everything on the right side is under the radical. plz helpp!

Answer 1

log₂(x-2) = √[log₂(x-2)]

Let u = log₂(x-2)

Then, u = √u

Squaring both sides yields the equation u² - u = 0

So, u(u-1) = 0

u = 0 or u = 1

log₂(x-2) = 0 or log₂(x-2) = 1

x-2 = 2⁰ or x - 2 = 2¹

^{x = 3 or x = 4}

Answer 2

log(2)x-2=√(log(2)x-2

this is true only when log(2)x-2=0 or 1 (0 equals its square root and 1 equals its square root)

so set log(2)x-2=0 and set log(2)x-2=1

log(2)x-2=0

log(2)x=2

2^2=x

x=4

log(2)x-2=1

log(2)x=3

2^3=x

x=8

if you don't see this do the following:

square both sides

log(2)x^2-4log(2)x+4=log(2)x-2

log(2)x^2-5log(2)x+6=0

let y=log(2)x

y^2-5y+6=0

use the quadratic equation to get y=2 and y=3 or factor to get (y-2)(y-3)=0

log(2)x=2

2^2=x

x=4

log(2)x=3

2^3=x

x=8

you see that you get 4 and 8 again