document.write( "Question 79561: Can anyone tell me the correct answer to this prob?? log(x-9)+log 100x = 3??
\n" ); document.write( "I keep getting .12 but when I plug it back into the problem it doesn't work. Thanks so much for your help. \n" ); document.write( "

Algebra.Com's Answer #57112 by Earlsdon(6294) $\"\"$ $\"About$

You can put this solution on YOUR website!
Solve for x:
\n" ); document.write( " $\"Log%5B10%5D%28x-9%29%2BLog%5B10%5D%28100x%29+=+3\"$ Apply the product rule for logarithms: $\"Log%5Bb%5D%28A%29%2BLog%5Bb%5D%28B%29+=+Log%5Bb%5D%28A%2AB%29\"$
\n" ); document.write( " $\"Log%5B10%5D%28%28x-9%29%2A%28100x%29%29+=+3\"$ Simplifying this, we get:
\n" ); document.write( " $\"Log%5B10%5D%28100x%5E2-900x%29+=+3\"$ Now write this in exponential form: $\"Log%5Bb%5D%28x%29+=+y\"$ means $\"b%5Ey+=+x\"$ Applying this to your problem:
\n" ); document.write( " $\"10%5E3+=+100x%5E2-900x\"$
\n" ); document.write( " $\"1000+=+100x%5E2-900x\"$ Subtract 1000 from both sides.
\n" ); document.write( " $\"100x%5E2-900x-1000+=+0\"$ Divide both sides by 100 to simplify this a bit.
\n" ); document.write( " $\"x%5E2-9x-10+=+0\"$ Factor this quadratic equation.
\n" ); document.write( " $\"%28x-10%29%28x%2B1%29+=+0\"$ Apply the principle of zero products:
\n" ); document.write( " $\"x-10+=+0\"$ or $\"x%2B1+=+0\"$ , so...
\n" ); document.write( " $\"x+=+10\"$ or $\"x+=+-1\"$ Discard the negative solution as the log of a negative quantity results in a complex answer.
\n" ); document.write( "Your solution is:
\n" ); document.write( " $\"x+=+10\"$
\n" ); document.write( "Check:
\n" ); document.write( " $\"Log%2810-9%29%2BLog%2810%2A100%29+=+Log%281%29%2BLog%281000%29\"$ = $\"0%2B3+=+3\"$ \n" ); document.write( "

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