document.write( "Question 237006: Solve for x: If log8 (x – 6) + log8 (x + 6) = 2, then x = \n" ); document.write( "
Algebra.Com's Answer #174347 by jsmallt9(3758)\"\" \"About 
You can put this solution on YOUR website!
You have a logarithmic equation with the variable in the argument(s). To solve such equations you want to transform the equation into one of the following forms:
\n" ); document.write( "log(some-expression-with-a-variable) = some-other-expression
\n" ); document.write( "or
\n" ); document.write( "log(some-expression-with-a-variable) = log(some-other-expression)
\n" ); document.write( "In other words, we want the logarithms to be the only thing on that side of the equation.

\n" ); document.write( "Since your equation has 3 terms, the 2 logs and 2, there is no way using simple Algebra to isolate both logs. But we have the properties of logarithms we can use. The one we need, \"log%28a%2C+%28p%29%29+%2B+log%28a%2C+%28q%29%29+=+log%28a%2C+%28p%2Aq%29%29\", allows us to combine the sum of two logs into a single log. Using this on your equation we get:
\n" ); document.write( "\"log%288%2C+%28%28x-6%29%28x%2B6%29%29%29+=+2\"
\n" ); document.write( "And in one step we have transformed the equation into the first of the two desired forms. From here, we rewrite the equation in exponential form. This will allow us to \"extract\" the variable from the log(s). \"log%28a%2C+%28p%29%29+=+b\" can be rewritten as \"a%5Eb+=+p\". Using this on your equation we get:
\n" ); document.write( "\"8%5E2+=+%28x-6%29%28x%2B6%29\"
\n" ); document.write( "Simplifying we get:
\n" ); document.write( "\"64+=+x%5E2+-+36\"
\n" ); document.write( "We have a quadratic equation. So we'll get one side equal to zero:
\n" ); document.write( "\"0+=+x%5E2+-100\"
\n" ); document.write( "Factor:
\n" ); document.write( "\"0+=+%28x%2B10%29%28x-10%29\"
\n" ); document.write( "And, since this (or any) product can be zero only if one of the factors is zero:
\n" ); document.write( "x+10 = 0 or x-10=0
\n" ); document.write( "Solving these we get:
\n" ); document.write( "x = -10 or x = 10

\n" ); document.write( "And finally, with logarithmic equations, we need to check our answers.
\n" ); document.write( "Checking x = -10:
\n" ); document.write( "\"log%288%2C+%28-10+-+6%29%29+%2B+log%288%2C+%28-10+%2B+6%29%29+=+2\"
\n" ); document.write( "We must reject -10 as a solution because it makes the arguments of the logs negative and arguments of logs can never be negative. (Note: Even if only one argument turned out negative we would still have to reject the solution.)

\n" ); document.write( "Checking x = 10:
\n" ); document.write( "\"log%288%2C+%2810+-+6%29%29+%2B+log%288%2C+%2810+%2B+6%29%29+=+2\"
\n" ); document.write( "\"log%288%2C+%284%29%29+%2B+log%288%2C+%2816%29%29+=+2\"
\n" ); document.write( "Using the property to combine the logs:
\n" ); document.write( "\"log%288%2C+%284%2A16%29%29+=+2\"
\n" ); document.write( "\"log%288%2C+%2864%29%29+=+2\"
\n" ); document.write( "Rewriting in exponential form:
\n" ); document.write( "\"8%5E2+=+64\"
\n" ); document.write( "Simplifying:
\n" ); document.write( "\"64+=+64\" Check!!

\n" ); document.write( "So there is only one solution: x = 10. \n" ); document.write( "
\n" );