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

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.

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

• Thank you! That was fast. I will accept your answer as soon as it will let me. – user500664 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 $b$? – user500664 Feb 25 '19 at 15:55
• @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

Alternatively,

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

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

Hence $$k = \log_3 n$$

• Thank you very much. This is even more elegant. I'm afraid I already accepted an answer – user500664 Feb 25 '19 at 16:12
• 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