How to proceed in this problem ?
If $ x = \log_{12} 27 \text {,then what is the the value of } \log_6 16 $?
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algebra-precalculus
logarithms
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1$\log_6 16=4\frac{\log\;2}{\log\;3+\log\;2}$ and $\log_{12} 27=3\frac{\log\;3}{\log\;3+2\log\;2}$ ... make of it what you will. – 2010-11-23
2 Answers
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$x = \log_{12}(27) = 3 \log_{12}(3)$.
$\frac{1}{x} = \frac{1}{3} \frac{1}{\log_{12}(3)} = \frac{1}{3} \log_{3}(12) = \frac{1}{3} \log_{3}(3 \times 4) = \frac{1}{3} (1 + 2 \log_{3}(2))$
Simplifying, we get $\log_{3}(2) = \frac{3}{2x} - \frac{1}{2}$.
Let $y = \log_{6}(16) = 4 \log_{6}(2) = 4 \frac{1}{\log_{2}(6)} = \frac{4}{1 + \log_{2}(3)}$.
Now make use of the fact that $\log_{3}(2) = \frac{1}{\log_{2}(3)}$ to get $y$ in terms of $x$.
I get $y = 4 (\frac{3-x}{3+x})$
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0Which gives us $y = 4 \cdot \biggl(\frac{3-x}{3+x}\biggr)$ – 2010-11-23
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Put $\rm\ \ell\: n\ =\ \log_6 n\:.\ \rm\ 27\ =\ 12^x\ \Rightarrow\ \ell\: 27\ =\ x\:(\ell\:6 + \ell\: 2)\ = x\:(1 + \ell\:2)\:.\ $ Now solve this for $\rm\:\ell\:2\:$ and plug the result into $\rm \ell\:16\ =\ 4\ \ell\: 2\ =\ \ldots$
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0Above $\rm\ell\:n$ is defined to be $\rm\log_6 n$. So e.g. $\ell\:12 = \ell\:(6\cdot 2) = \ell\:6 + \ell\:2 = 1 + \ell\: 2$ – 2010-11-23