# If #log_12 27 = a#, then what is #log_6 16#?

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If ( \log_{12} 27 = a ), then ( \log_{6} 16 ) can be found using the change of base formula for logarithms:

[ \log_{b} x = \frac{\log_{a} x}{\log_{a} b} ]

So,

[ \log_{6} 16 = \frac{\log_{12} 16}{\log_{12} 6} ]

Using the properties of logarithms, we know that ( \log_{12} 16 = \frac{\log_{12} 2^4}{\log_{12} 6} ).

Solving ( \log_{12} 2^4 ), we get ( 4 \log_{12} 2 ).

Now, putting it all together:

[ \log_{6} 16 = \frac{4 \log_{12} 2}{\log_{12} 6} = \frac{4}{a} ]

So, ( \log_{6} 16 = \frac{4}{a} ).

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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.

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.

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.

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