Trigonometry Question

Question

8log₂ √(3x - 2) - log₂( 4/x) + log₂4

Answer 1

First notice log₂(4/x) = log₂4 - log₂x

so we can write 8log₂(√3x-2)-log₂(4/x)+log₂4 = 8log₂(√3x-2) - log₂4 + log₂x + log₂4 = 8log₂(√3x-2) + log₂x

Now 8log₂(√3x-2) = log₂((√3x-2)⁸) = log₂((3x-2)⁴)

So we need to combine log₂((3x-2)⁴) + log₂(x). I'll let you figure this out. Remember log₂(x)+log₂(y)=log₂(xy)