log 3 (12) = log 3 (3x +27)

We know that:

if log a b = log a c

==> b = c

In our case, we could wrire:

12 = 3x + 27

Subtract 27 from both sides:

-15 = 3x

Now divide by3:

==> -5 = x

==> x= -5

Now let us check the answer:

log 3 12 = log 3 (3(-5)+27)

log 3 12 = log 3 (-15+27)

log 3 12 = log 3 (12)

Then the answer is:

x= -5

Before solving the equation, we have to impose constraints of existance of logarithm function.

3x+27>0

We'll add -27 both sides:

3x>-27

We'll divide by 3:

x>-9

So, for the logarithms to exist, the values of x have to be in the interval (-9, +inf.)

Because the bases of the logarithms are matching, we'll use the one to one property:

3x+27= 12

3x = 12-27

3x = -15

We'll divide by 3:

**x = -5 > -9**

The solution is admissible because the value belongs to the interval (-9,+inf.)

To solve x. log3 (12) = log3 (3x+27)

We lnow that m log a = log a^m. So the given equation becomes:

log3 (12) = = log 3 (3x+27). Taking antilog to base 3,we get:

12= 3x+27

3x = 12-27 = -15.

x = -15/3 = -5.