document.write( "Question 286068: log2(5-x)+log2(5+x)=4 \n" ); document.write( "

Algebra.Com's Answer #207426 by Earlsdon(6294) $\"\"$ $\"About$

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Solve for x:
\n" ); document.write( " $\"Log%5B2%5D%285-x%29%2BLog%5B2%5D%285%2Bx%29+=+4\"$ Apply the \"product rule\" for logarithms: $\"highlight_green%28Log%5Bb%5D%28M%29%2BLog%5Bb%5D%28N%29+=+Log%5Bb%5D%28M%2AN%29%29\"$
\n" ); document.write( " $\"Log%5B2%5D%28%285-x%29%285%2Bx%29%29+=+4\"$ Simplify the left side.
\n" ); document.write( " $\"Log%5B2%5D%2825-x%5E2%29+=+4\"$ Recall the definition of the logarithm:
\n" ); document.write( "\"The logarithm of a number (or an expression) is the power to which the base must be raised to equal that number (or expression)\"
\n" ); document.write( "Rewrite the above equation accordingly:
\n" ); document.write( " $\"2%5E4+=+25-x%5E2\"$ Simplify.
\n" ); document.write( " $\"16+=+25-x%5E2\"$ Rearrange into a standard quadratic equation.
\n" ); document.write( " $\"x%5E2-9+=+0\"$ Factor the left side.
\n" ); document.write( " $\"%28x%2B3%29%28x-3%29+=+0\"$ Apply the \"zero product rule\"
\n" ); document.write( " $\"x%2B3+=+0\"$ or $\"x-3+=+0\"$ therefore:
\n" ); document.write( " $\"x+=+-3\"$ or $\"x+=+3\"$ \n" ); document.write( "

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