004 Sample Final A, Problem 15

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

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

Solution:

Step 1:
Using a rule of logarithms, the equation becomes $\log((x+8)(x-1))=1$.

Step 2:
By the definition of the logarithm, $\log((x+8)(x-1))=1$
means $10=(x+8)(x-1)$

Step 3:
Now, we can solve for $x$. We have $0=(x+8)(x-1)-10=x^{2}+7x-18=(x+9)(x-2)$.
So, there are two possible answers, which are $x=-9$ or $x=2$.

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

Final Answer:
$x=2$

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