document.write( "Question 436590: solve for x: 2log2x-log21=log29 \n" ); document.write( "
Algebra.Com's Answer #302403 by lwsshak3(11628)\"\" \"About 
You can put this solution on YOUR website!
solve for x: 2log2x-log21=log29
\n" ); document.write( "..
\n" ); document.write( "2log2x-log21=log29
\n" ); document.write( "2log2x-log21-log29=0
\n" ); document.write( "2log2x-(log21+log29)=0
\n" ); document.write( "log((2x)^2)/(21*29)=0
\n" ); document.write( "convert to exponential form: (base raised to log of number=number) In this case base=10, log of the number=0, and the number=((2x)^2)/(21*29)
\n" ); document.write( "10^0=4x^2/609=1
\n" ); document.write( "4x^2=609
\n" ); document.write( "x^2=609/4
\n" ); document.write( "x=+-sqrt(609/4)=12.33896
\n" ); document.write( "x=12.33896
\n" ); document.write( "x=-12.33896 (reject, 2x>0) \n" ); document.write( "
\n" );