You need to move the term` -log(3x)` to the right side, such that:

`log(x + 7) = log(3x) + log (5)`

Converting the summation `log(3x)` + `log (5)` into the logarithm of product yields:

`log(x + 7) =log(5*3x)`

`log(x + 7) = log(15x) => x + 7 = 15x => x - 15x = -7 => -14x = -7 => x = 7/14 => x = 1/2`

Testing the value `x = 1/2` in logarithms yields:

`log(1/2 + 7) = log(15*1/2)`

`log((1+14)/2) = log(15/2)`

`log(15/2) = log(15/2)`

**Hence, testing the value `x = 1/2` in equation, it holds, hence, evaluating the solution to the given equation yields `x = 1/2` .**

We'll write the equation

log (x+7) - log 3x = log 5

We could use the quotient property of the logarithms:

log a - log b = log (a/b)

We'll put a = x+7 and b = 3x and we'll get:

log [(x+7)/3x] = log 5

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

(x+7)/3x = 5

We'll cross multiply:

x+7 = 15x

We'll isolate x to the left side. For this reason, we'll subtract 15 both sides:

-14x = -7

We'll divide by -14 both sides:

x = 7/14

x = 1/2

Since the value of x is positive, the solution of the equation is admissible and it is x = 1/2.