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Given that #log_b(N^4)=8#, what is #log_b(1/N)#?

Answer 1

Ariana Alvarez

#-2#

#log_b(N^4)=4log_bN=8#. So, #log_bN=2#. And so,

#log_b(1/N)=log_b(N^(-1))=-log_bN=-2#

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Answer 2

Elijah Allen

To find log_b(1/N), we can use the properties of logarithms.

log_b(N^4) = 8

Using the power rule of logarithms, we can rewrite this equation as:

4 * log_b(N) = 8

Now, let's solve for log_b(N):

log_b(N) = 8 / 4 log_b(N) = 2

Now, to find log_b(1/N), we can use the property of logarithms that states log_b(1/N) = -log_b(N).

So,

log_b(1/N) = -log_b(N)

Substituting the value of log_b(N) we found earlier:

log_b(1/N) = -2

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.