SOLUTION: 4^log base 2 of 3

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Question 711567: 4^log base 2 of 3
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Expressions of the form:
a%5Elog%28a%2C+%28p%29%29
are very simple to simplify. Your expression
4%5Elog%282%2C+%283%29%29
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:
%282%5E2%29%5Elog%282%2C+%283%29%29
At this point we have a power of a power. The rule for the exponents for this is to multiply the exponents:
2%5E%282%2Alog%282%2C+%283%29%29%29

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:
2%5Elog%282%2C+%283%5E2%29%29
which simplifies to:
2%5Elog%282%2C+%289%29%29
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,
log%282%2C+%289%29%29
is the exponent one would put on a 2 to get a result of 9. And look at where
log%282%2C+%289%29%29
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
2%5Elog%282%2C+%289%29%29 must be a 9!

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:
    a%5Elog%28a%2C+%28p%29%29+=+x
    Find the base a log of each side:
    log%28a%2C+%28a%5Elog%28a%2C+%28p%29%29%29%29+=+log%28a%2C+%28x%29%29
    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.
    log%28a%2C+%28p%29%29%2Alog%28a%2C+%28a%29%29+=+log%28a%2C+%28x%29%29
    We should know that log%28a%2C+%28a%29%29+=+1 so this simplifies to:
    log%28a%2C+%28p%29%29+=+log%28a%2C+%28x%29%29
    The equation now says that two base a logs are equal. If so then the arguments must be equal. So:
    p=x
    Since p+=+x and a%5Elog%28a%2C+%28p%29%29+=+x:
    a%5Elog%28a%2C+%28p%29%29+=+p