Question

Solve \[\log_{3} \][(x+5)(x-3)]=2

Answer 1

In exponential form this is equivalent to solving \((x+5)(x-3) = 3^2 \implies x^2+2x-24 = 0\).

Can you take things from here?

Answer 2

I am supposed to come up with a solution set

Answer 3

Right...so what do you get when you solve that quadratic equation?

Answer 4

its \[\log_{3} \] in front of it

Answer 5

Did you include the log3?

Answer 6

You can rewrite logarithmic equations as exponential equations. In this case, we have that
\[\log_3((x+5)(x-3)) = 2 \implies (x+5)(x-3)=3^2\implies x^2+2x-24 = 0.\]

Answer 7

Oh! I understand now. So it should be: (x+6) (x-4)

Answer 8

x=-6, x=4 ?

Answer 9

Okay, that's good so far, so now you need to test them in the original equation to see if any of your solutions are extraneous.

Answer 10

In this case, it doesn't seem like we get any extraneous solutions, so that would imply that both of the values you found are your solutions. Does this make sense? :-)

Answer 11

Yes! Thank you