# 008A Sample Final A, Question 16

Question: Solve. ${\displaystyle \log _{6}(x+2)+\log _{6}(x-3)=1}$

Foundations:
1) How do we combine the two logs?
2) How do we remove the logs?
1) One of the rules of logarithms says that ${\displaystyle \log(x)+\log(y)=\log(xy)}$
2) The definition of logarithm tells us that if ${\displaystyle \log _{6}(x)=y}$, then ${\displaystyle 6^{y}=x}$

Solution:

Step 1:
Using one of the properties of logarithms the, left hand side is equal to ${\displaystyle \log _{6}((x+2)(x-3)}$
Step 2:
By the definition of logarithms ${\displaystyle \log _{6}((x+2)(x-3)=1}$ means ${\displaystyle 6=(x+2)(x-3)}$
Step 3:
Now we do some arithmetic to solve for x. ${\displaystyle 0=(x+2)(x-3)-6=x^{2}-x-12=(x-4)(x+3)}$. So there are two possible answers.
Step 4:
We have to make sure the answers make sense in the context of the problem. Since the domain of the log function is ${\displaystyle (0,\infty )}$  ,   -3 is removed as a potential answer.