We have : log(3x+5) - log(5-x) = log 2

log(3x+5) - log(5-x) = log 2

use the property of logarithm log a - log b = log (a/b)

=> log[(3x + 5)/(5 - x) = log 2

=> (3x + 5)/(5 - x) = 2

=> 3x + 5 = 10 - 2x

=> 5x = 5

=> x = 1

**The solution of log(3x+5) - log(5-x) = log 2 is x = 1**

First, we'll impose the necessary conditions of existence of logarithms:

3x+5>0 => 3x>-5 => x > -5/3

and

5-x>0 => x<5

The common interval of admissible values is (-5/3 ; 5).

We'll use the quotient property of logarithms:

log (3x+5)/(5-x) = log 2

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

(3x+5)/(5-x) = 2

We'll multiply by (5-x) both sides:

3x + 5 = 10 - 2x

We'll add 2x both sides:

5x = 5

x = 1

**Since the value 1 belongs to the interval of admissible values for x, the equation has the solution x = 1.**