document.write( "Question 255028: I'm just wondering if it's possible to simplify this further or if I'm even on the right track...\r
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document.write( "3^x=15 (find x)\r
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document.write( ">x=log₃15
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document.write( ">x=log₃(3x5)
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document.write( ">x=log₃3 + log₃5
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document.write( ">x= 1 + log₃5\r
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document.write( "Thanks for your help :o) \n" );
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Algebra.Com's Answer #187364 by Alan3354(69443)![]() ![]() You can put this solution on YOUR website! 3^x=15 (find x) \n" ); document.write( ">x=log₃15 \n" ); document.write( ">x=log₃(3x5) \n" ); document.write( ">x=log₃3 + log₃5 \n" ); document.write( ">x= 1 + log₃5 \n" ); document.write( "= 1 + (log(5)/log(3)) \n" ); document.write( "---------------------- \n" ); document.write( "Second line should be \n" ); document.write( "x log3 = log 15 giving \n" ); document.write( "x = log 15 / log 3 = 1.1770 / 0.4771 \n" ); document.write( "------------------ \n" ); document.write( "The result is the same, but Greenfinch's method is faster and simpler. \n" ); document.write( "= 2.46497... \n" ); document.write( " |