008A Sample Final A, Question 16

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

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

Solution:

Step 1:
Using one of the properties of logarithms the, left hand side is equal to $\log _{6}((x+2)(x-3)$

Step 2:
By the definition of logarithms Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \log _{6}((x+2)(x-3)=1} means $6=(x+2)(x-3)$

Step 3:
Now we do some arithmetic to solve for x. Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 $(0,\infty )$  ,   -3 is removed as a potential answer.

Final Answer:
x = 4.

Return to Sample Exam