Logarithmic equation 1

Question

$2\log(x-1)-\log2=\log(x+7)$

$\log(x-1)^2 -\log2=\log(x+7)$

$\log \frac{(x-1)^2}{2}=\log(x+7)$

$\frac{(x-1)^2}{2}=x+7$

$(x-1)^2=2x+14$

$x^2-4x-13=0$

$\frac{Δ}{4}=17$

Have I made some mistake?

Answer 1

HINT

Note from the next-to-last line that $$x^2-4x-13 = (x-2)^2-17$$

Answer 2

Nope. Your solution seems to be correct.

But do ensure that the values you get for $x$ after solving satisfies the domain on question, which is $x>1$.