SOLUTION: log5(x+2)+log5(x-2)=3

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Question 238168: log5(x+2)+log5(x-2)=3
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
When you are solving for variables that are in the argument(s) of logarithms, you want to use Algebra and the properties of logarithms to transform the equation into the form:
log(expression-with-variables) = some-other-expression

Transforming your equation is not difficult. All we need to do is use the property, log%28a%2C+%28p%29%29+%2B+log%28a%2C+%28q%29%29+=+log%28a%2C+%28p%2Aq%29%29. This allows us to "combine" the sum of these two different logs into one. Your equation, after using this property, now looks like:
log%285%2C+%28%28x%2B2%29%28x-2%29%29%29+=+3
And we have the desired form. Now we will rewrite this in exponential form, using the fact that log%28a%2C+%28p%29%29+=+q is equivalent to p+=+a%5Eq:
%28x%2B2%29%28x-2%29+=+5%5E3
Simplifying this we get:
x%5E2+-+4+=+125
Now we can add 4 to both sides:
x%5E2+=+129
Square root of both sides:
sqrt%28x%5E2%29+=+sqrt%28129%29
abs%28x%29+=+sqrt%28129%29
Solving this we get:
x+=+sqrt%28129%29 or x+=+-sqrt%28129%29

With logarithm equations it is important to check your answers. (It's more than just a good idea.) What we check for is to make sure that none of the arguments of the logarithms become zero or negative.
Checking x+=+sqrt%28129%29
log%285%2C+%28sqrt%28129%29%2B2%29%29+%2B+log%285%2C+%28sqrt%28129%29-2%29%29+=+3
Since sqrt%28129%29+%3E+2 both arguments are positive.

Checking x+=+-sqrt%28129%29
log%285%2C+%28-sqrt%28129%29%2B2%29%29+%2B+log%285%2C+%28-sqrt%28129%29-2%29%29+=+3
Both arguments are negative so we have to reject this solution. (Important: It was not because -sqrt%28129%29 was negative that we rejected it. We rejected it because the logarithm arguments became negative when x was -sqrt%28129%29 that we rejected it. In other problems it might work out that a negative value for x would be OK but a positive one would not. And sometimes both answers check and sometimes no answers work.)