8A F11 Q16

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 a rule of logarithms the left hand side is equal to $\log _{6}((x+2)(x-3)$

Step 2:
By the definition of logarithms $\log _{6}((x+2)(x-3)=1$ means $6=(x+2)(x-3)$

Step 3:
Now we do some arithmetic to solve for x. $0=(x+2)(x-3)-6=x^{2}-x-12=(x-4)(x+3)$. 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.

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