Solving $\left(1/3\right)^k n = 1$ for $k$

Question

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.

Answer 1

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

Answer 2

Alternatively,

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

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

Hence $$k = \log_3 n$$