Logarithms are a weak point for me and I'm curious how my professor went from the following logarithm to the next one. How are they equal? $$ 3^{\log_4 n} = n^{\log_4 3} $$
And does that mean I can change $2^{\log_2{n}}$ to $n^{\log_2{2}}$?
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See also: https://math.stackexchange.com/q/2891737 https://math.stackexchange.com/q/320116 https://math.stackexchange.com/q/1626298 https://math.stackexchange.com/q/3518266 – Martin Sleziak Jan 22 '20 at 08:41
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$x^y=e^{y\ln x}$ and $\log_{x}y = \frac{\ln y}{\ln x}$. Hence $$3^{\log_4n}=e^{\ln 3\frac{\ln n}{\ln 4}}=e^{\ln n\frac{\ln 3}{\ln 4}}=n^{\log_43}$$
Your proposed change is fine, although you don't really need it; $2^{\log_2n}=n$, since the exponentiation and logarithm cancel, being to the same base.
vadim123
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Well, the change he made makes that easy to see: $2^{\log_2 n}=n^{\log_2 2}=n^1=n$. – mjqxxxx Oct 22 '13 at 14:24
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I disagree; $2\times \frac{1}{2}\times x=x$ is just as easy to see as $x\times 2\times \frac{1}{2}=x$. – vadim123 Oct 22 '13 at 14:26