How can I simplify this so I don't have a log in the exponent ? $3n - 3 * 2^{\log _{3}(n)}$
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Maybe this will help: $2 = 3^{\log_{3}(2)}$, and so $2^{\log _{3}(n)} = 3^{\log_{3}(2)\log_{3}(n)} = n^{\log_{3}(2)}$, so in total:
$$ 3n - 3 \cdot 2^{\log _{3}(n)} = 3n - 3n^{\log_{3}(2)} $$
Wolfgang Kais
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Call $3n - 3 * 2^{\log _{3}(n)} = y $
'Exponentiate' both sides by 3,
You end up with $ 3^y = \frac {3^{3n}}{3^{(3)(2)^{\log _{3}(n)}}}$
Can you go from there?
Going further,
$ 3^y = \frac {3^{3n}}{3^{(3)(3^{ \log_3 (2) \log_3 (n))}}} = \frac {3^{3n}}{3^{3n^{ \log_3 (2)}}}$
$ \ln(3^y) = 3n \ln(3) - 3n^{ \log_3 (2)} ln(3) = y \ln(3)$
$ y = 3n - 3n^{ \log_3 (2)}$
khaled014z
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Thank you very much ! – Shanks Feb 05 '19 at 00:18
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Your calculation contains a mistake: $(3)(2)^{\log_{3}(n)} = \log_{3}(n^6)$ (which is $(3)(2){\log_{3}(n)}$) is true if and only if $n \in {3, 9}$. – Wolfgang Kais Feb 06 '19 at 14:18
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Thank you, I edited it, it became a more complicated version of your answer though. – khaled014z Feb 06 '19 at 17:24