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The goal is to show that $$\left(\frac{1}{3}\right)^kn=1 \Rightarrow k = \log_3 n\,.$$

So I started with $\left(\frac{1}{3}\right)^kn=1 \Leftrightarrow \left(\frac{1}{3}\right)^k=\frac{1}{n}$ in order to use the identity $y=a^x \Leftrightarrow x=\log_a y$, which then yields $$k=\log_{1/3} \frac{1}{n}$$ which using $\log \frac{1}{x}=-\log a$ can be written as $$k = -\log_{1/3} n\,.$$ But that is not what I wanted to show, as $\log_3 n \neq -\log_{1/3} n$.

I don't know where the mistake is.

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Note that $$-\log_{1/3} n = \frac{\log_{1/3} n}{\log_{1/3}3} = \log_3 n$$

Haris Gušić
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    Thank you! That was fast. I will accept your answer as soon as it will let me. –  Feb 25 '19 at 15:45
  • Sorry, I've got another question: the identity you used is $\log_b (a)=\frac{\log_c a}{\log_c b}$, right? In your answer, $c=\frac{1}{3}$, but couldn't it be any arbitrary number as $c$ bears no relation to either $a$ or $b$? –  Feb 25 '19 at 15:55
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    @ThomasFlinkow Yes, but keep in mind that $0<c \neq 1$ for the logarithms to be defined. – Haris Gušić Feb 25 '19 at 15:58
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Alternatively,

$$\left( \frac{1}{3^k}\right)n=1$$

Multiplying $3^k$ on both sides, $$n=3^k$$

Hence $$k = \log_3 n$$

Siong Thye Goh
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  • Thank you very much. This is even more elegant. I'm afraid I already accepted an answer –  Feb 25 '19 at 16:12
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    Don't worry about reputations. I just see a question and just lift a few fingersl ;) The other solution teach you why $-\log_{\frac13} n = \log_3 n$. – Siong Thye Goh Feb 25 '19 at 16:16