document.write( "Question 481432: please help me prove that-
\n" ); document.write( " $\"16%5Elog3+=+9%5Elog4\"$ \n" ); document.write( "

Algebra.Com's Answer #329653 by kingme18(98) $\"\"$ $\"About$

You can put this solution on YOUR website!
First, note that both bases are perfect squares: $\"16=4%5E2\"$ and $\"9=3%5E2\"$ . Let's rewrite your equation:
\n" ); document.write( " $\"16%5Elog3+=+9%5Elog4\"$
\n" ); document.write( " $\"%284%5E2%29%5Elog3=%283%5E2%29%5Elog4\"$
\n" ); document.write( "Recall that when you raise a power to a power, you multiply the exponents:
\n" ); document.write( " $\"4%5E%282%2Alog3%29=3%5E%282%2Alog4%29\"$
\n" ); document.write( "We are allowed to take the logarithm of both sides of the equation:
\n" ); document.write( " $\"log4%5E%282%2Alog3%29=log3%5E%282%2Alog4%29\"$
\n" ); document.write( "My favorite property of logarithms is that you can \"kick the exponent to the front\" :)
\n" ); document.write( " $\"2%2Alog3%2Alog4=2%2Alog4%2Alog3\"$
\n" ); document.write( "Rearrange to make it obvious (commutative property):
\n" ); document.write( " $\"2%2Alog3%2Alog4=2%2Alog3%2Alog4\"$ \n" ); document.write( "

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