Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How to proceed in this problem ?

share|improve this question
    
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 $x$? –  Arturo Magidin Nov 23 '10 at 5:12
    
Yes, we are supposed to express $\log_6 16$ in terms of $x$. –  Quixotic Nov 23 '10 at 5:14
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. –  J. M. Nov 23 '10 at 5:15

2 Answers 2

up vote 5 down vote accepted

$x = \log_{12}(27) = 3 \log_{12}(3)$.

$\frac{1}{x} = \frac{1}{3} \frac{1}{\log_{12}(3)} = \frac{1}{3} \log_{3}(12) = \frac{1}{3} \log_{3}(3 \times 4) = \frac{1}{3} (1 + 2 \log_{3}(2))$

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 $y = 4 (\frac{3-x}{3+x})$

share|improve this answer
    
Which gives us $y = 4 \cdot \biggl(\frac{3-x}{3+x}\biggr)$ –  Quixotic Nov 23 '10 at 5:37

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 $\rm \ell\:16\ =\ 4\ \ell\: 2\ =\ \ldots$

share|improve this answer
    
Anybody,care to explain what does this mean in simple terms ? –  Quixotic Nov 23 '10 at 5:54
    
@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. –  Bill Dubuque Nov 23 '10 at 5:57
    
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 :( –  Quixotic Nov 23 '10 at 5:59
    
@Deb: Where are you stuck? –  Bill Dubuque Nov 23 '10 at 6:01
    
Ok, for this post, I think $\ell n$ is Napierian Logarithm,but am not able to figure out the rest. –  Quixotic Nov 23 '10 at 6:05

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.