document.write( "Question 115701: log base5 (3x+15)- log base 5 of 3x=1 \n" ); document.write( "
Algebra.Com's Answer #84196 by solver91311(24713)\"\" \"About 
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\"log%285%2C3x%2B15%29-log%285%2C3x%29=1\"\r
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\n" ); document.write( "\n" ); document.write( "You need to use three of the rules for logarithms for this problem.\r
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\n" ); document.write( "\n" ); document.write( "1) \"log%28b%2Cx%5Ea%29=a%2Alog%28b%2Cx%29\", and
\n" ); document.write( "2) \"log%28b%2Cf%28x%29%29%2Blog%28b%2Cg%28x%29%29=log%28b%2Cf%28x%29%2Ag%28x%29%29\"
\n" ); document.write( "3) \"log%28b%2Cx%29=y\" is equivalent to \"b%5Ey=x\"\r
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\n" ); document.write( "\n" ); document.write( "Using the first rule, you can write:\r
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\n" ); document.write( "\n" ); document.write( "\"log%285%2C3x%2B15%29%2Blog%285%2C3x%5E%28-1%29%29=1\", notice that the minus sign in the original equation became an exponent of -1.\r
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\n" ); document.write( "\n" ); document.write( "\"log%285%2C3x%2B15%29%2Blog%285%2C%281%2F3x%29%29=1\" (using the rule \"x%5E%28-1%29=1%2Fx\")\r
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\n" ); document.write( "\n" ); document.write( "Now applying rule 2) we can write:
\n" ); document.write( "\"log%285%2C%283x%2B15%29%2A%281%2F3x%29%29=1\"\r
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\n" ); document.write( "\n" ); document.write( "Now, using rule 3) we can write:
\n" ); document.write( "\"5%5E1=%283x%2B15%29%2A%281%2F3x%29\"\r
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\n" ); document.write( "\n" ); document.write( "Simplify and solve for x
\n" ); document.write( "\"15x=3x%2B15\"
\n" ); document.write( "\"12x=15\"
\n" ); document.write( "\"4x=5\"
\n" ); document.write( "\"x=5%2F4\". Done.\r
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