If $\log_{2n} 1994 = \log_n \left(486 \sqrt{2}\right)$, compute $n^6$.
Problem
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Tags: logarithms
Silverfalcon
28.01.2006 01:15
In original test, the problem never told you this but to give your answer in exact form, write your answer in form $p_1^{e_1} \cdot p_2^{e_2} \cdot p_3^{e_3} \cdots$ where $p_i$ is the prime and $e_i$ is the corresponding exponent with $i = 1,2,3,...$
TachyonPulse
02.06.2009 02:00
This is actually a simple problem, but it is impossible to have an integer value of $ n$ with the given values. $ 1944$ was the number used in the original problem and it turns out that it works out very nicely.
Factorizing the two integers in the problem, we obtain $ 486 = 2\cdot 3^5$ and $ 1944 = 2^3\cdot 3^5$.
Let $ a = \log_{2n}{1944} = \log_n{486\sqrt{2}}$. We have
$ 2^a\cdot n^a = 2^3 \cdot 3^5 (1)$ and $ n^a = 2\cdot 3^5\cdot 2^{\frac{1}{2}}\implies n^a = 2^{\frac{3}{2}}\cdot 3^5 (2)$.
Substituting $ (2)$ into $ (1)$, we have
$ 2^a \cdot 2^{\frac{3}{2}}\cdot 3^5 = 2^3 \cdot 3^5\implies 2^a = 2^{\frac{3}{2}}\implies a = \frac{3}{2}$.
Finally, using $ (2)$, we have
$ n^{\frac{3}{2}} = 2^{\frac{3}{2}}\cdot 3^5\implies n^6 = \boxed{2^6\cdot 3^{20}}$.