SOLUTION: Anybody out there who can help with eveluating expressions?? Ineed to evaluate the following expression; log g 1? Thanks a Million.

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Question 105567This question is from textbook Intermediate Algebra
: Anybody out there who can help with eveluating expressions?? Ineed to evaluate the following expression; log g 1?
Thanks a Million. This question is from textbook Intermediate Algebra

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Given:
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log%28g%2C1%29=+y
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This means that y is the value of log%28g%2C1%29
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The answer to this problem is that y is equal to zero, so you can say that log%28g%2C1%29
is equal to zero. Notice that you are saying that the log of 1 is equal to zero regardless
of what the base is.
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Let's see why. We can do this by translating from the logarithmic form to its equivalent
exponential form. This translation says that:
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log%28b%2CA%29=y has an equivalent exponential form b%5Ey+=+A
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By comparing the log form of the translation to the log equation you were given in this
problem, you can see that:
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b+=+g
A+=+1 and
y+=+y
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Then you can see that by substituting the right side of these three relationships into
the exponential form, that the exponential form of the given problem is:
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g%5Ey+=+1
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Now it's time to recall that if any base number is raised to the zero power, the answer is 1.
Since we need 1 as the answer, we need to raise g to the zero power. So y = 0 and this
means that log%28g%2C1%29 equals zero.
.
Hope this helps you to understand this problem and how to get the answer.
.