document.write( "Question 31386: Solve the equation: \r
\n" ); document.write( "\n" ); document.write( "log(3+x)-log(x-3)=log 3 \n" ); document.write( "

Algebra.Com's Answer #18093 by Earlsdon(6294) $\"\"$ $\"About$

You can put this solution on YOUR website!
To solve this, you would use the \"quotient\" rule for logarithms rather than the \"product\" rule for logarithms.
\n" ); document.write( "The quotient rules states:
\n" ); document.write( " $\"log%28M%29+-+log%28N%29+=+log%28%28M%2FN%29%29\"$ Applying this rule to your problem, we have:
\n" ); document.write( " $\"log%28%283%2Bx%29%29+-+log%28%28x-3%29%29+=+log%28%28%283%2Bx%29%2F%28x-3%29%29%29\"$ and this = $\"log%28%283%29%29\"$ , so:
\n" ); document.write( " $\"log%28%28%283%2Bx%29%2F%28x-3%29%29%29+=+log%28%283%29%29\"$ Therefore:
\n" ); document.write( " $\"%28%283%2Bx%29%29%2F%28%28x-3%29%29+=+3\"$ Now you can solve for x. Multiply both sides by (x-3)
\n" ); document.write( " $\"3%2Bx+=+3%28x-3%29\"$ Simplify.
\n" ); document.write( " $\"3%2Bx+=+3x-9\"$ Subtract x from both sides.
\n" ); document.write( " $\"3+=+2x-9\"$ Add 9 to both sides.
\n" ); document.write( " $\"12+=+2x\"$ Finally, divide both sides by 2.
\n" ); document.write( " $\"6+=+x\"$ \r
\n" ); document.write( "\n" ); document.write( "Solution is:
\n" ); document.write( "x = 6\r
\n" ); document.write( "\n" ); document.write( "Check:
\n" ); document.write( " $\"log%28%283%2B6%29%29-log%28%286-3%29%29+=+log%28%289%29%29-log%28%283%29%29\"$ Applying the \"quotient\" rule, we get:
\n" ); document.write( " $\"log%28%289%29%29-log%28%283%29%29+=+log%28%289%2F3%29%29\"$ = $\"log%28%283%29%29\"$ \n" ); document.write( "

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