document.write( "Question 55179: HELP ME PLEASE:
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document.write( "SOLVE LOGa(8x+5)=LOGa(4x+29) \n" );
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Algebra.Com's Answer #37437 by funmath(2933)![]() ![]() ![]() You can put this solution on YOUR website! SOLVE LOGa(8x+5)=LOGa(4x+29) \n" ); document.write( "Because the bases are the same, we can simply equate: \n" ); document.write( "8x+5=4x+29 \n" ); document.write( "-4x+8x+5=-4x+4x+29 \n" ); document.write( "4x+5=29 \n" ); document.write( "4x+5-5=29-5 \n" ); document.write( "4x=24 \n" ); document.write( "4x/4=24/6 \n" ); document.write( "x=6 \n" ); document.write( "check by substituting x=6 in the original equation. You have to do that with all log equations to avoid extraneous (false) solutions. \n" ); document.write( "LOGa(8(6)+5)=LOGa(4(6)+29) \n" ); document.write( "LOGa(48+5)=LOGa(24+29) \n" ); document.write( "LOGa(53)=LOGa(53) We're right!!! \n" ); document.write( "Happy Calculating!!! \n" ); document.write( " |