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I'm stuck on continuing the next exercise:

Considering:

$\log_{c}a = 3$
$\log_{c}b = 4$

and:

$ y = \frac{a^{3}\sqrt{b \cdot c^{2}}}{2} $

What's the value of $\log_{c}y$ (integer)?

So far, I did all the substitutions that were obvious at my eyes:

$ y = \frac{a^{3}\sqrt{b \cdot c^{2}}}{2}\quad\rightarrow\quad y = \frac{c^{9}\sqrt{c^{4} \cdot c^{2}}}{2}\quad\rightarrow\quad y = \frac{c^{9}\sqrt{c^{6}}}{2}\quad\rightarrow\quad y = \frac{c^{9} c^{3}}{2}\quad\rightarrow\quad y = \frac{c^{12}}{2} $

The last equality is the same as $c^{12} = 2y$, which could be written as:

$ \log_{c}2y = 12\quad\rightarrow\quad \log_{c}2 + \log_{c}y = 12 $

I really don't know haw to continue. I've managed to find a $\log_{c}y$, but I don't know what to do with the $\log_{c}2$. Either I took the wrong path, or I'm missing something that prevents me to finish this exercise.

Any hints are much welcome! Thanks in advance.

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    Yeah, assuming that the course is something along the lines of advanced algebra, precalculus, or college algebra, it's almost certainly a misprint.2011-05-12

2 Answers 2

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Take $\log_c$ of both sides of the equation:

$ \log_c y=3\log_c a+\frac12\log_c b+1-\log_c 2.$ And so, $\log_c y=3(3)+\frac12(4)+1-\log_c 2=12-\log_c 2.$

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    A lot easier… Thanks for pointing it out Nana.2011-05-12
3

Note that we know $\log_c y$ is an integer, so $\log_c 2$ must also be an integer, hence $c^k=2$ for some $k\in\mathbb{Z}$. If we also assume $c$ is an integer, then the problem does in fact have a unique solution. Hope this helps.

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    Very clever! Well, following the idea $\log_{c}2 = 1$ (if I'm not mistaken), but unfortunately I've just checked the workbook answers (answers are provided, resolutions not) and it should be 4, not 11. Hmm, probably there is something wrong with the exercise. I'll search for my teacher tomorrow in school and ask him about this exercise. Thanks a lot you all, I'll post any news. cc @Isaac @GEdgar @Nana2011-05-12