I'm stuck on continuing the next exercise:


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


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

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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    $\begingroup$ At a quick glance, your work looks right to me and I don't see a way to deal with the 2 or $\frac{1}{2}$. $\endgroup$ – Isaac May 11 '11 at 23:15
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    $\begingroup$ Perhaps $a,b,c$ are to be integers. After all, we don't have the source book... $\endgroup$ – GEdgar May 11 '11 at 23:58
  • $\begingroup$ @Isaac, thanks. I was also thinking that could be no apparent solution, but know @GEdgar and @Skatche proposals are clarifying some aspects. The source book is really vague. I'll check the answer — I guess there is an error with this exercise. $\endgroup$ – sidyll May 12 '11 at 0:04
  • $\begingroup$ Yeah, assuming that the course is something along the lines of advanced algebra, precalculus, or college algebra, it's almost certainly a misprint. $\endgroup$ – Isaac May 12 '11 at 0:18

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

  • $\begingroup$ A lot easier… Thanks for pointing it out Nana. $\endgroup$ – sidyll May 12 '11 at 0:10

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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    $\begingroup$ 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 @Nana $\endgroup$ – sidyll May 12 '11 at 0:15

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