log x + log 2x = 3 log 2 find x values

We'll start by imposing constraints of existence of
logarithms:

x>0

2x>0

The
interval of admissible values of x is (0 , +infinite).

Now,
we'll solve the equation. We'll start by applying the power property of
logarithms:

3 log 2 = log 2^3 = log
8

Now, we'll subtract log 2x both
sides:

log x = log 8 - log
2x

We'll apply the quotient rule of
logarithms:

log a - log b = log
a/b

log x = log 8/2x

log x =
log 4/x

Since the bases are matching, we'll apply the one
to one property:

x = 4/x

x^2 =
4

x1 = +sqrt4

x1 =
+2

x2 = -2

We'll reject the
second solution because it doesn't belong to the interval of admissible
values.

The only solution of the equation is
x = 2.