document.write( "Question 186596: log3 + 2 = log x\r
\n" );
document.write( "\n" );
document.write( "solve for x\r
\n" );
document.write( "\n" );
document.write( "logs are to base 10 \n" );
document.write( "
Algebra.Com's Answer #139858 by stanbon(75887) ![]() You can put this solution on YOUR website! log3 + 2 = log x \n" ); document.write( "------------------- \n" ); document.write( "log(x) - log(3) = 2\r \n" ); document.write( "\n" ); document.write( "log(x/3) = 2\r \n" ); document.write( "\n" ); document.write( "x/3 = 10^2\r \n" ); document.write( "\n" ); document.write( "x/3 = 100\r \n" ); document.write( "\n" ); document.write( "x = 300 \n" ); document.write( "============= \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( " |