How do I prove or disprove that $n^{\log_2 c} = c^{\log_2 n}$

Question

I have tried it for a couple hours now, closest I got was:

$c = 2^{\log_2 c}$

$n = 2^{\log_2 n}$

then: $$2^{\log_2 n \log_2 c} = c^{\log_2 n}$$

Can you help me progress?

Answer 1

$2^{\log_2 n \log_2 c} = c^{\log_2 n}\implies$

$2^{(\log_2 n)\times (\log_2 c)} = c^{\log_2 n}\implies$

$[2^{\log_2 n}]^{\log_2 c} = c^{\log_2 n}\implies$

$n^{\log_2 c} = c^{\log_2 n}$

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Alternatively let $2^k =n$ and $2^m = c$ so $\log_2 n =k$ and $\log_2 c = m$.

Then $n^{\log_2 c} = n^m$ And $c^{\log_2 n} = c^k$.

And $n^m = (2^k)^m= 2^{km}$ and $c^k = (2^m)^k = 2^{mk}$.

So to be thorough: $n^{\log_2 c}= n^m = (2^k)^m = 2^{km}=(2^m)^k=c^k=c^{\log_2 n}$.

Or to be efficient as spare:

$n^{\log_2 c} =2^{\log_2 n\cdot \log_2 c}=2^{\log_2 c\cdot \log_2 n} =c^{\log_2 n}$.

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Or if you want to be a smart-ass....

$n^{\log_2 c} = c^{\log_2 n} \iff $

$\log_2 (n^{\log_2 c}) = \log_2(c^{\log_2 n}) \iff$

$\log_2 c\cdot \log_2 n = \log_2 n \cdot \log_2 c$

Which it does.

Answer 2

$$n^{\log_2c}=c^{\log_cn\log_2c}=c^{\frac{\log_cn}{\log_c2}}=c^{\log_2n}$$

Answer 3

You're almost there. You can show that $n^{\log_2 c}$ also equals $2^{\log_2 n \log_2 c}$.

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Alternatively, applying $\log_2$ to both sides of the desired equality yields $$\log_2 (c) \log_2(n) = \log_2(n) \log_2(c)$$

Answer 4

You can take the logarithm of both sides to obtain $$\log_{2}{(c)}\log{n}=\log_{2}{(n)}\log{c},$$ which you can verify is true by the change of base formula.