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How to proceed in this problem ?

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    Are you sure you've got the right logarithms? Are you expected to express $\log_{6}(16)$ using $x$ in some way? If not, what's the point of $x$?2010-11-23
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    Yes, we are supposed to express $\log_6 16$ in terms of $x$.2010-11-23
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    $\log_6 16=4\frac{\log\;2}{\log\;3+\log\;2}$ and $\log_{12} 27=3\frac{\log\;3}{\log\;3+2\log\;2}$ ... make of it what you will.2010-11-23

2 Answers 2

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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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    Which 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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    Anybody,care to explain what does this mean in simple terms ?2010-11-23
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    @Deb: As always, if you tell me what is not clear I am happy to elaborate. But don't give up so easily either. An important part of learning how to solve math problems is figuring out how to make the "leaps" between various hints along the way.2010-11-23
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    Weird but as like,almost most of the times,I am not getting anything out of it ...I despise giving up, but I don't know why I get mired every-time ... I try to figure out your hints :(2010-11-23
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    @Deb: Where are you stuck?2010-11-23
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    Ok, for this post, I think $\ell n$ is Napierian Logarithm,but am not able to figure out the rest.2010-11-23
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    Above $\rm\ell\:n$ 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