document.write( "Question 275957: Can you help me simplify $\"%28125%5E%282log%285%2C+x%29%29%29%283%5E%28log%289%2C+%28x%5E10%29%29%29%29\"$ ? I believe it's $\"x%5E21\"$ , but I'm not sure. Thanks! \n" ); document.write( "

Algebra.Com's Answer #201275 by jsmallt9(3758) $\"\"$ $\"About$

You can put this solution on YOUR website!
$\"%28125%5E%282log%285%2C+x%29%29%29%283%5E%28log%289%2C+%28x%5E10%29%29%29%29\"$
\n" ); document.write( "The key to this problem is understanding that by definition:
\n" ); document.write( " $\"a%5Elog%28a%2C+%28p%29%29+=+p\"$
\n" ); document.write( "So we want to make the exponent of 125 a base 125 logarithm and to make the exponent of 3 a base 3 logarithm. There is a formula for changing bases, $\"log%28a%2C+%28p%29%29+=+log%28b%2C+%28p%29%29%2Flog%28b%2C+%28a%29%29\"$ , but we can't use it if the logarithm has a (visible) coefficient. So we need to start by using a property of logarithms, $\"q%2Alog%28a%2C+%28p%29%29+=+log%28a%2C+%28p%5Eq%29%29\"$ , which allows us to move a coefficient into the argument of the logarithm as its exponent. We will use this to move the 2 into the argument of the logarithm:
\n" ); document.write( " $\"%28125%5E%28log%285%2C+%28x%5E2%29%29%29%29%283%5E%28log%289%2C+%28x%5E10%29%29%29%29\"$
\n" ); document.write( "Now we can use the change of base formula to change the two exponents into base 125 and base 3 logarithms, respectively:
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\n" ); document.write( "Since the cube root of 125 is 5 ans since cube root is an exponent of 1/3, $\"125%5E%281%2F3%29+=+5\"$ and $\"log%28125%2C+%285%29%29+=+1%2F3\"$ . And $\"log%283%2C+%289%29%29+=+2\"$ . Substituting both of these into the denominators of the exponents we get:
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\n" ); document.write( "Changing the divisions in the exponents into multiplying by the reciprocals we get:
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\n" ); document.write( "Using our property to move coefficients into arguments again we get:
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\n" ); document.write( "which simplifies to:
\n" ); document.write( " $\"%28125%5E%28log%28125%2C+%28x%5E6%29%29%29%29%283%5E%28log%283%2C+%28x%5E5%29%29%29%29\"$
\n" ); document.write( "By definition of logarithms
\n" ); document.write( " $\"125%5E%28log%28125%2C+%28x%5E6%29%29%29+=+x%5E6\"$ and $\"3%5E%28log%283%2C+%28x%5E5%29%29%29+=+x%5E5\"$ . Substituting we get:
\n" ); document.write( " $\"%28x%5E6%29%2A%28x%5E5%29\"$
\n" ); document.write( "which simplifies to:
\n" ); document.write( " $\"x%5E11\"$ \n" ); document.write( "

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