# 004 Sample Final A, Problem 15

Solve. ${\displaystyle \log(x+8)+\log(x-1)=1}$

Foundations
1) How can we combine the two logs?
2) How do we remove logs from an equation?
1) One of the rules of logarithms states that ${\displaystyle \log(x)+\log(y)=\log(xy)}$
2) The definition of the logarithm tells us that if ${\displaystyle \log(x)=y}$, then ${\displaystyle 10^{y}=x}$.

Solution:

Step 1:
Using a rule of logarithms, the equation becomes ${\displaystyle \log((x+8)(x-1))=1}$.
Step 2:
By the definition of the logarithm, ${\displaystyle \log((x+8)(x-1))=1}$
means ${\displaystyle 10=(x+8)(x-1)}$
Step 3:
Now, we can solve for ${\displaystyle x}$. We have ${\displaystyle 0=(x+8)(x-1)-10=x^{2}+7x-18=(x+9)(x-2)}$.
So, there are two possible answers, which are ${\displaystyle x=-9}$ or ${\displaystyle x=2}$.
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

${\displaystyle (0,\infty )}$, -9 is removed as a potential answer. The answer is ${\displaystyle x=2}$.

${\displaystyle x=2}$