document.write( "Question 202237: Solve for x: log(2x-3)=logx+log(x-2) thanks. \n" ); document.write( "

Algebra.Com's Answer #152520 by Earlsdon(6294) $\"\"$ $\"About$

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Solve for x:
\n" ); document.write( " $\"Log%282x-3%29+=+Log%28x%29%2BLog%28x-2%29\"$ Apply the \"product rule\" for logarithms to the right side of the equation. $\"Log%5Bb%5D%28M%2AN%29+=+Log%5Bb%5D%28M%29%2BLog%5Bb%5D%28N%29\"$
\n" ); document.write( " $\"Log%282x-3%29+=+Log%28%28x%29%2A%28x-2%29%29\"$ Simplify the argument on the right side.
\n" ); document.write( " $\"Log%282x-3%29+=+Log%28x%5E2-2x%29\"$ Apply the \"identity\" property.:If $\"Log%5Bb%5D%28M%29+=+Log%5Bb%5D%28N%29\"$ then $\"M+=+N\"$
\n" ); document.write( " $\"2x-3+=+x%5E2-2x\"$ Form into a standard quadratic equation.
\n" ); document.write( " $\"x%5E2-4x%2B3+=+0\"$ Factor.
\n" ); document.write( " $\"%28x-1%29%28x-3%29+=+0\"$ Apply the \"zero product\" rule.
\n" ); document.write( " $\"x-1+=+0\"$ or $\"x-3+=+0\"$ therefore:
\n" ); document.write( " $\"highlight%28x+=+1%29\"$ or $\"highlight%28x+=+3%29\"$ \n" ); document.write( "

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