We could subtract log 5 both sides:

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

We could use the quotient property of the logarithms:

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

We'll use the one to one property:

3x = (x+7)/5

We'll cross multiply:

15x = x+7

We'll isolate x to the left side:

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.

To solve log3x+log5 = log(x+7)

Solution:

log3x+log5 = log(x+7).

This could be written as:

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

3x*5 = (x+7) , by one to one property of logaritm function with the same base.

15x =x+7.

15x-x =7.

14x = 7.

x = 7/14 or

x = 1/2.

Given:

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

Using the formula:

log a + log b = log (a*b)

The given expression can be converted into:

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

==> log 15x = log (x + 7)

Therefore:

15x = x + 7

==> 15x - x = 7

==> 14x = 7

Therefore:

x = 7/14 = 1/2 = 0.5