# Solve for x log 3 (2x+3) -1 = 0.

*print*Print*list*Cite

### 3 Answers

log 3 (2x+3) -1 = 0

Move 1 to the right side:

==> log 3 (2x+3) = 1

We know that log 3 (3) = 1

==> log 3 (2x + 3) = log 3 (3)

We know that:

if log a = log b ==> a = b

==> 2x + 3 = 3

Subtract 3:

==> 2x = 0

==> **x= 0**

We'll start by imposing the constraints of existance of logarithm function.

2x+3>0

We'll add -3 both sides:

2x>-3

We'll divide by 2:

x>-3/2

So, for the logarithms to exist, the values of x have to belong to the interval (-3/2, +inf.)

We'll shift the free term to the right side:

log 3 (2x+3) = 1

We'll create matching bases to the right side.

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

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

2x+3 = 3

We'll eliminate like terms:

2x = 3-3

2x = 0

We'll divide by 2:

**x = 0 > -3/2**

Since the value for x belongs to the interval (-3/2,+inf.), the solution is valid.

log(2x+3)-1 = 0. To solve for x.

Solution:

log3 (2x+3) -1 = 0. Add 1.

log3 (2x+3) = 1.

log3 (2x+3) = log 3 (3). Take antilog:

(2x+3) = 3

2x = 0

x = 0