document.write( "Question 711567: 4^log base 2 of 3 \n" ); document.write( "
Algebra.Com's Answer #437522 by jsmallt9(3758)\"\" \"About 
You can put this solution on YOUR website!
Expressions of the form:
\n" ); document.write( "\"a%5Elog%28a%2C+%28p%29%29\"
\n" ); document.write( "are very simple to simplify. Your expression
\n" ); document.write( "\"4%5Elog%282%2C+%283%29%29\"
\n" ); document.write( "is not quite in that form because the base of the exponent, 4, is different from the base of the logarithm, 2. But here is where we can figure out a relatively simple solution. Aren't 4 and 2 powers of each other? (\"4+=+2%5E2\" and \"2+=+4%5E%281%2F2%29\"). So we should be able to convert one base, the 4 or the 2, into the other. I'm going to change the base of 4 into a base of 2. Replacing the 4 with \"2%5E2\" we get:
\n" ); document.write( "\"%282%5E2%29%5Elog%282%2C+%283%29%29\"
\n" ); document.write( "At this point we have a power of a power. The rule for the exponents for this is to multiply the exponents:
\n" ); document.write( "\"2%5E%282%2Alog%282%2C+%283%29%29%29\"

\n" ); document.write( "We have the bases matching. But we are still not quite in the desired form. The 2 in front of the log is the problem. But fortunately there is a property of logarithms, \"n%2Alog%28a%2C+%28p%29%29+=+log%28a%2C+%28p%5En%29%29\", which allows us to move a coefficient in front of the log into the argument of the log as its exponent. Using this property we can move the 2 (which is in our way) into the argument as its exponent:
\n" ); document.write( "\"2%5Elog%282%2C+%283%5E2%29%29\"
\n" ); document.write( "which simplifies to:
\n" ); document.write( "\"2%5Elog%282%2C+%289%29%29\"
\n" ); document.write( "We now have the desired form. If you don't yet know how this simplifies, let's review what logarithms are. Logarithms in general are exponents. Base 2 logarithms are exponents one would put on a 2. More specifically,
\n" ); document.write( "\"log%282%2C+%289%29%29\"
\n" ); document.write( "is the exponent one would put on a 2 to get a result of 9. And look at where
\n" ); document.write( "\"log%282%2C+%289%29%29\"
\n" ); document.write( "is in our expression. It is the exponent on a 2! So our expression is \"2 to the power that you would put on a 2 to get a result of 9\". I hope it is clear that this means that
\n" ); document.write( "\"2%5Elog%282%2C+%289%29%29\" must be a 9!

\n" ); document.write( "If this is not clear then either:
  • Memorize: \"a%5Elog%28a%2C+%28p%29%29+=+p\"; or
  • Learn to simplify it as follows. Let's say that:
    \n" ); document.write( "\"a%5Elog%28a%2C+%28p%29%29+=+x\"
    \n" ); document.write( "Find the base a log of each side:
    \n" ); document.write( "\"log%28a%2C+%28a%5Elog%28a%2C+%28p%29%29%29%29+=+log%28a%2C+%28x%29%29\"
    \n" ); document.write( "Use the property of logarithms mentioned earlier. This time, however, we are going to in the other direction: move the exponent of the argument out in front.
    \n" ); document.write( "\"log%28a%2C+%28p%29%29%2Alog%28a%2C+%28a%29%29+=+log%28a%2C+%28x%29%29\"
    \n" ); document.write( "We should know that \"log%28a%2C+%28a%29%29+=+1\" so this simplifies to:
    \n" ); document.write( "\"log%28a%2C+%28p%29%29+=+log%28a%2C+%28x%29%29\"
    \n" ); document.write( "The equation now says that two base a logs are equal. If so then the arguments must be equal. So:
    \n" ); document.write( "\"p=x\"
    \n" ); document.write( "Since \"p+=+x\" and \"a%5Elog%28a%2C+%28p%29%29+=+x\":
    \n" ); document.write( "\"a%5Elog%28a%2C+%28p%29%29+=+p\"
\n" ); document.write( "
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