SOLUTION: Having trouble with a logarithm! I am okay with logarithms, but this one throws in a square root so I am lost as to what to do with it. log base-2 (log base-b (sqrt b)) , where

Question 379858: Having trouble with a logarithm! I am okay with logarithms, but this one throws in a square root so I am lost as to what to do with it.
log base-2 (log base-b (sqrt b)) , where b > 0.
I just had a thought.. do I need to turn sqrt b into b^(1/2)?
Any help would be appreciated, thanks! :)
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!

You are correct. start by rewriting as :

Next, on the inner logarithm, we can use a property of logarithms, , to "move" the exponent of the argument out in front of the logarithm:

By definition, so this becomes:

The quick way to finish requires that we understand what this expression represents. It represents the exponent for 2 that results in 1/2. If you also understand exponents, you will know that the answer is: -1!

The long way is to use another property of logarithms, , to split the numerator and denominator into separate logarithms:

The logarithm of 1, regardless of the base, is zero (because any number to the zero power is 1). And by definition . So now we have:
0 - 1
which equals
-1
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