document.write( "Question 946643: Solve fora in log(2a)- 3log2= 1/2log(a-3) \n" ); document.write( "

Algebra.Com's Answer #854629 by MathTherapy(10858) $\"\"$ $\"About$

You can put this solution on YOUR website!

\r\n" );
document.write( "Solve fora in log(2a)- 3log2= 1/2log(a-3)\r\n" );
document.write( "*****************************************\r\n" );
document.write( "The SMALLEST log argument of the 3 log arguments, log (a - 3), signifies that a - 3 > 0, and so, a MUST BE > 3.\r\n" );
document.write( "     , with a being > 3.\r\n" );
document.write( "  2 log (2a) - 6 log (2) = log (a - 3) ----- Multiplying by LCD, 2\r\n" );
document.write( "2 log (2a) - log (a - 3) = 6 log (2) --- Subtracting log (a - 3) and adding 6 log (2), to both sides\r\n" );
document.write( "   ---- Applying  = \r\n" );
document.write( " \r\n" );
document.write( "                ----- Applying  = \r\n" );
document.write( "                           ---- If , then c = d\r\n" );
document.write( "                              ---- Cross-multiplying\r\n" );
document.write( "                                ---- Dividing each side by 4\r\n" );
document.write( "                               \r\n" );
document.write( "              \r\n" );
document.write( "            (a - 12)(a - 4) = 0 --- Factorizing TRINOMIAL\r\n" );
document.write( "                             \r\n" );
document.write( "\r\n" );
document.write( "a = 12 > 3, and a = 4 > 3. Therefore, 12 and 4 are VALID/ACCEPTABLE values of a.

\n" ); document.write( "

\n" );