How is $2^{\log_4 n}= n^{\log _42}$?
I saw in a notebook the following: $2^{\log_4 n}= n^{\log _42}(=\sqrt n)$, but I never saw this before and I can't find it in any log rules, is it right? and if so how did they do it?
BTW, if we take $\log_2$ of both sides we get: $${\log_4 n} = {\log _42}\log_2 n= \frac 1 2 \log_2 n$$ which makes both sides look even more different.
EDIT: OK, never mind. I found that rule:
Funny they don't mention it everywhere...
$\endgroup$ 27 Answers
$\begingroup$Take the logarithm in base $4$: \begin{align} &\mathrm{(LHS)} & \log_4(2^{\log_4n})=(\log_4n)(\log_42) \\ &\mathrm{(RHS)} & \log_4(n^{\log_42})=(\log_42)(\log_4n) \end{align} Since the logarithms are equal, the numbers are equal.
$\endgroup$ $\begingroup$Notice, log rule: $\color{red}{\log_{a^n}(b)=\frac 1n\log_a(b)}$ hence, simplifying LHS & RHS as follows
$$LHS=2^{\log_4n}=2^{\log_{2^2}n}=2^{\frac{1}{2}\log_2n}=2^{\log_2n^{1/2}}=n^{1/2}=\color{blue}{\sqrt n}$$ $$RHS=n^{\log_42}=n^{\log_{2^2}2}=n^{\frac{1}{2}\log_22}=n^{\frac 12}=\color{blue}{\sqrt n}$$
$\endgroup$ $\begingroup$Hint: Try rewriting $$2^{\log_4 n}$$ as $$\bigg(4^{\frac{1}{2}}\bigg)^{\log_4 n}$$
$\endgroup$ $\begingroup$They aren't different. $\log_4(2)=\frac12$ because $4^{1/2}=2$.
$\endgroup$ $\begingroup$The left side is equivalent to $$2^{\frac{log(n)}{log(4)}}$$ and the right side is equivalent to $$n^{\frac{log(2)}{log(4)}}$$
The equation is therefore true, when $2^{log(n)}=n^{log(2)}$
Logarithming both sides gives $log(n)log(2)=log(2)log(n)$, so the given equation is actually true. Because of $\frac{log(2)}{log(4)}=\frac{1}{2}$, the result is $\sqrt{n}$ in both cases.
$\endgroup$ $\begingroup$...or use the property of the natural logarithm:
$$ t = 2^{\log_2 t} = 2^{\frac{\ln t}{\ln 2}} $$
$\endgroup$ $\begingroup$In general we have:
$\forall a > 1, \ \forall b > 1, \ \log_a n = \log_a b \ \cdot \ \log_b n$
Proof:
$\log_a n = \frac{\ln n}{\ln a} = \frac{\ln n}{\ln a} \ \cdot \ \frac{\ln b}{\ln b} = \frac{\ln b}{\ln a} \ \cdot \ \frac{\ln n}{\ln b} = \log_a b \ \cdot \ \log_b n$
Therefore we have:
$\mathrm{2}^{\log_4 n} = \mathrm{2}^{\log_4 2 \ \cdot \ \log_2 n} = \mathrm{2}^{\log_2 n \ \cdot \ \log_4 2} = n^{\log_4 2}$
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