You need to convert the sum of logarithms into the logarithm of product, such that:

`log x + log (2x) = 3log 2 => log (x*2x) = 3 log 2`

`log (2x^2) = log 8 => 2x^2 = 8 => x^2 = 8/2 => x^2 = 4`

`x^2 - 4 = 0`

Converting the difference of squares `x^2 - 4` into a product, yields:

`{(x^2 - 4 = (x - 2)(x + 2)),(x^2 - 4 = 0):} => (x - 2)(x + 2) = 0 => {(x - 2 = 0),(x + 2 = 0):} => {(x = 2),(x = -2):}`

Testing the values `x = 2` and `x = -2` in equation, yields:

`x = 2 => log (2*2*2) = 3 log 2 => log 8 = log 8`

`x = -2 => log (-2) + log (-4) = log 8` invalid

**Hence, evaluating the solutions to the given equation, yields `x = 2` .**

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.

The equation log x+log 2x = 3log2 has to be solved.

Use the following properties of logarithms: log a + log b = log(a*b), a*log b = log b^a

log x+log 2x = 3log2

log(x*2x) = log 2^3

log 2x^2 = log 8

Equate 2x^2 and 8 and solve for x

2x^2 = 8

x^2 = 4

x = 2

The given equation has only one solution and that is x = 2