I ran across this simple identity yesterday, but can’t seem to find a way to get from one side to the other: $$ n^{2} = 4^{{\log_{2}}(n)}. $$
Wolfram Alpha tells me that it is true, but other than that, I’m stuck.
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2By the way, in mathematics you just use "$=$", not "$==$", to say that two things are equal. It's called the equals sign after all. – Feb 07 '13 at 07:51
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$$4^{\log_2n}=\left(2^2\right)^{\log_2n}=2^{2\log_2n}=\left(2^{\log_2n}\right)^2=n^2$$
Brian M. Scott
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The outside exponent in the second to last term should be a $2$, not an $n$. – Aaron Feb 07 '13 at 07:34
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Recall that $$a^{\log_a(x)} = x \,\,\,\,\,\, \forall a,x >0$$ Hence, $$4^{\log_2(n)} = (2^{2})^{\log_2(n)} = (2^{\log_2(n)})^2 = n^2$$
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Oh I'm an idiot, I forgot that apart from $(2^m)^n == 2^{m*n}$ we could also swap the m and n.. obvious now that someone tells me. – Voo Feb 07 '13 at 07:35
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Take $\log_{2}$ of both sides and get $$n^{2} = 4^{{\log_{2}} n}$$ $$2\log_{2}n =2\log_{2}n $$
user 1591719
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This is the natural way taught in high school for such questions. (Sister) – user 1591719 Feb 07 '13 at 10:49
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Let $n=2^t$ then for LHS we have : $n^2=(2^t)^2=2^{2t}$
and for RHS : $4^{Log_2 n}=4^{Log_2 2^t}=4^t=(2^2)^t=2^{2t}$
LHS = RHS $\square$
jimjim
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