SOLUTION: log base of 9 (x^2+12x+17)=2

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Question 274273: log base of 9 (x^2+12x+17)=2
Answer by jsmallt9(3296) About Me  (Show Source):
You can put this solution on YOUR website!
log%289%2C+%28x%5E2%2B12x%2B17%29%29=2
To solve a logarithmic equation where the variable is in the base or the argument of a logarithm, you start by getting the equation into one of the following forms:
log(variable-expression) = other-expression
or
log(variable-expression) = log(other-expression)

Fortunately your equation is already in the first form. With this form we proceed by rewriting the equation in exponential form:
x%5E2%2B12x%2B17+=+9%5E2
which simplifies to:
x%5E2%2B12x%2B17+=+81
Now we solve this equation. Since it is a quadratic equation, we'll get one side equal to zero (by subtracting 81 from each side):
x%5E2%2B12x-64+=+0
Then we factor it (or use the Quadratic Formula). This factors fairly easily:
%28x%2B16%29%28x-4%29+=+0
From the Zero Product Property we know that this product is zero only if one of the factors is zero. So:
x%2B16+=+0 or x-4+=+0
Solving these we get:
x+=+-16 or x+=+4

When solving logarithmic equations it is important (not just a good idea) to check your answers. We must make sure that no bases or arguments of logarithms are zero or negative. Always use the original equation to check:
log%289%2C+%28x%5E2%2B12x%2B17%29%29=2
Checking x = -16:
log%289%2C+%28%28-16%29%5E2%2B12%28-16%29%2B17%29%29=2
which simplifies to:
log%289%2C+%28256%2B12%28-16%29%2B17%29%29=2
log%289%2C+%28256%2B+%28-192%29+%2B+17%29%29=2
log%289%2C+%2881%29%29=2
And since 9%5E2+=+81, log%289%2C+%2881%29%29+=+2 Check!

Checking x = 4:
log%289%2C+%28%284%29%5E2%2B12%284%29%2B17%29%29=2
which simplifies to:
log%289%2C+%2816%2B12%284%29%2B17%29%29=2
log%289%2C+%2816%2B+48+%2B+17%29%29=2
log%289%2C+%2881%29%29=2
And since 9%5E2+=+81, log%289%2C+%2881%29%29+=+2 Check!

Both solutions check out. If either or both solutions had made the base or the argument of a logarithm zero or negative we would have to reject it/them.