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

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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.**

The equation `log_3 (x+3)+ log_3 (x-5)=2` has to be solved.

Use the property of logarithm `log_b a + log_b c = log_b(a*c)` and if` log_b a = c` , `a = b^c`

`log_3 (x+3)+ log_3 (x-5)=2`

=> `log_3 ((x+3)/(x - 5)) = 2`

=> `(x+3)/(x-5) = 3^2 = 9`

=> x + 3 = 9x - 45

=> 8x = 48

=> x = 48/8 = 6

The solution of the equation is x = 6