# What is x if log(3x+5)-log(5-x)=log 2?

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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.**

## 3x+5/5-x=2 (loga-logb=loga/logb)

therefore ** 2(**5-x)=3x+5

10-2x=3x+5

5x=5 ; x=1