document.write( "Question 55179: HELP ME PLEASE:
\n" ); document.write( "SOLVE LOGa(8x+5)=LOGa(4x+29)
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Algebra.Com's Answer #37437 by funmath(2933)\"\" \"About 
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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!!!
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