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)![]() ![]() ![]() You can put this solution on YOUR website! 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 \n" ); document.write( " |