document.write( "Question 825920: Please i need help solving for x, thank you
\n" ); document.write( "log(x-1)+log(1-x)=2 \n" ); document.write( "

Algebra.Com's Answer #497703 by josgarithmetic(39620) $\"\"$ $\"About$

You can put this solution on YOUR website!
Jumping ahead a couple of steps, let b = the base of the logarithm ( you did not indicate the base, so I might assume you mean base ten), $\"b%5E2=%28x-1%29%281-x%29\"$ ;\r
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\n" ); document.write( "\n" ); document.write( "If base ten logs, then $\"%28x-1%29%28x-1%29%28-1%29=100\"$
\n" ); document.write( " $\"x%5E2-2x%2B1=100\"$
\n" ); document.write( " $\"x%5E2-2x-99=0\"$
\n" ); document.write( " $\"%28x-11%29%28x%2B9%29=0\"$
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\n" ); document.write( "x=-9 or x=11 \n" ); document.write( "

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