how to solve log v3 (x+3)+ log v3 (x-5)=2the "V" means it is a base

log3 ( x+ 3) + log3 ( x-5) =
2

First we will determine the
domain,

==> x+ 3 > 0   and x-5 >
0

==> x > -3   and   x >
5

==> x > 5 is the
domain.......(1)

We will use logarithm properties to solve
for x.

We know that log a + log b= = log
ab

==> log3 (x+3) + log 3 ( x-5) = log3 (x+3)*(x-5)
= 2

Now we will open the
brackets.

==> log3 ( x^2 - 2x - 15) =
2

Now we will rewrite using he exponent
form.

==>  x^2 -2x - 15 =
3^2

==> x^2 - 2x - 15 =
9

==> x^2 - 2x - 24 =
0

Now we will
factor.

==> (x -6)(x+4) =
0

==> x = 6 > 5  ( belongs to the
domain)

==> x = -4  < 5 ( does NOT belong to
the domain)

Then the only solution is x =
6.