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

Answer 1

Brooklyn Anderson

An excellent issue

Given

#a=log_12 27=log_(12)3^3=3log_(12)3# #=>a=3/log_3 12=3/log_3(3xx2^2)#

#=>a=3/(log_3 3+2log_3 2#

#=>a=3/(1+2log_3 2)#

#=>(1+2log_3 2)=3/a#

#=>log_3 2=1/2(3/a-1)=(3-a)/(2a)#

#=>log_2 3=(2a)/(3-a)#

Now #log_6 16=log_6 2^4=4log_6 2=4/log_2 6#

#=4/log_2 (2xx3)=4/(log_2 2+log_2 3#

#=4/(1+log_2 3)=4/(1+(2a)/(3-a)#

#=(4(3-a))/(3-a+2a)=(12-4a)/(3+a)#

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Answer 2

Josiah Anderson

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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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.