document.write( "Question 739759: log2(3x-5)>log2(x+7)\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "I'm so stumped. Any help would by appreciated. \n" ); document.write( "
Algebra.Com's Answer #451309 by Edwin McCravy(20060)\"\" \"About 
You can put this solution on YOUR website!
\r\n" );
document.write( "log2(3x - 5) > log2(x + 7)\r\n" );
document.write( "\r\n" );
document.write( "Since logarithms with base greater than 1 are always increasing,\r\n" );
document.write( "what they are logs of are in the same order of inequality as the\r\n" );
document.write( "logs themselves are.  Therefore we know that\r\n" );
document.write( "\r\n" );
document.write( "    3x - 5 > x + 7\r\n" );
document.write( "        2x > 12\r\n" );
document.write( "         x > 6 \r\n" );
document.write( "\r\n" );
document.write( "Edwin
\n" ); document.write( "
\n" );