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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.
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