document.write( "Question 193122: This is a problem I am having a hard time doing. log(x+9)=1-logx \n" ); document.write( "
Algebra.Com's Answer #144962 by RAY100(1637)\"\" \"About 
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log(x+9)=1-logx
\n" ); document.write( "log(x+9) +logx=1
\n" ); document.write( "to multiply is to add logs
\n" ); document.write( "log(x+9)*(x)=1
\n" ); document.write( "log(x^2+9x)=1
\n" ); document.write( "raise both sides as power of 10
\n" ); document.write( "10^log(x^2+9x)=10^1
\n" ); document.write( "(x^2+9x) =10
\n" ); document.write( "x^2+9x-10=0
\n" ); document.write( "(x+10)(x-1)=0
\n" ); document.write( "x=1, and -10 (not realistic answer, no log(- number)
\n" ); document.write( "check x=1
\n" ); document.write( "log(1+9)=1-log1
\n" ); document.write( "log10=1-log1
\n" ); document.write( "1=1-0 ok
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