document.write( "Question 1084208: Log6 54 + log6 4 = \n" ); document.write( "

Algebra.Com's Answer #698275 by jim_thompson5910(35256) $\"\"$ $\"About$

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\n" ); document.write( "Log Rules:
\n" ); document.write( "Let b be the base of the logarithm. The base must positive, so b > 0. Also we can't let b = 1 be possible so the restriction is that $\"b+%3C%3E1\"$ . For any positive x,y values, we can say
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$\"log%28b%2C%28x%29%29%2Blog%28b%2C%28y%29%29=log%28b%2C%28x%2Ay%29%29\"$
$\"log%28b%2C%28x%5Ey%29%29=y%2Alog%28b%2C%28x%29%29\"$
$\"log%28b%2C%28b%29%29=1\"$

We'll use these log rules below
\n" ); document.write( "-----------------------------------------\r
\n" ); document.write( "\n" ); document.write( " $\"log%286%2C%2854%29%29%2Blog%286%2C%284%29%29=log%286%2C%2854%2A4%29%29\"$ Using log rule 1
\n" ); document.write( " $\"log%286%2C%2854%29%29%2Blog%286%2C%284%29%29=log%286%2C%28216%29%29\"$
\n" ); document.write( " $\"log%286%2C%2854%29%29%2Blog%286%2C%284%29%29=log%286%2C%286%5E3%29%29\"$
\n" ); document.write( " $\"log%286%2C%2854%29%29%2Blog%286%2C%284%29%29=3%2Alog%286%2C%286%29%29\"$ Using log rule 2
\n" ); document.write( " $\"log%286%2C%2854%29%29%2Blog%286%2C%284%29%29=3%2A1\"$ Using log rule 3
\n" ); document.write( " $\"log%286%2C%2854%29%29%2Blog%286%2C%284%29%29=3\"$
\n" ); document.write( " \n" ); document.write( "

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