# Thread: Why does this equation hold

1. ## Why does this equation hold

Why does this equation hold true? I cannot seem to algebraically prove this.

$\displaystyle 2^{log n} = n^{log 2}$

thanks for the help in advance.

2. ## Re: Why does this equation hold

Originally Posted by bakerconspiracy
Why does this equation hold true? I cannot seem to algebraically prove this.

$\displaystyle 2^{log n} = n^{log 2}$

thanks for the help in advance.
It only holds true for the values of n such that n^2 = 2^n.

3. ## Re: Why does this equation hold

In my homework we have a solution that reduces from $\displaystyle 2^{lg^2(n) } = 2^{lg(n) * lg(n)} = n^{lg(n)}$ where $\displaystyle lg(n) = log_2(n)$. Can you explain how this happens then? Does it only work for the values you described?

Originally Posted by Prove It
It only holds true for the values of n such that n^2 = 2^n.

4. ## Re: Why does this equation hold

Originally Posted by bakerconspiracy
Why does this equation hold true? I cannot seem to algebraically prove this.

$\displaystyle 2^{log n} = n^{log 2}$

thanks for the help in advance.
I do apologise, I misread your question. I thought the LHS was \displaystyle \displaystyle \begin{align*} 2\log{(n)} \end{align*}, not \displaystyle \displaystyle \begin{align*} 2^{\log{(n)}} \end{align*}, and the RHS was \displaystyle \displaystyle \begin{align*} n\log{(2)} \end{align*}, not \displaystyle \displaystyle \begin{align*} n^{\log{(2)}} \end{align*}.

Anyway, the best thing to do is to convert them to the same base

\displaystyle \displaystyle \begin{align*} 2^{\log{(n)}} &\equiv e^{\log{\left[2^{\log{(n)}}\right]}} \\ &\equiv e^{\log{(n)}\log{(2)}} \\ &\equiv \left[e^{\log{(n)}}\right]^{\log{(2)}} \\ &\equiv n^{\log{(2)}} \end{align*}

Q.E.D.

5. ## Re: Why does this equation hold

Hello, bakerconspiracy!

$\displaystyle \text{Prove: }\:2^{\log n} = n^{\log 2}$

$\displaystyle \text{Let }X \:=\:2^{\log n}$

$\displaystyle \text{Take logs: }\;\log X \:=\:\log\left(2^{\log n}\right) \:=\:\log n\cdot\log 2 \:=\:\log 2\cdot\log n$

. . . . . . . . $\displaystyle \log X \;=\;\log\left(n^{\log 2}\right)$

$\displaystyle \text{Exponentiate: }\;X \;=\;n^{\log 2}$

$\displaystyle \text{Therefore: }\;2^{\log n} \;=\;n^{\log 2}$

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### equation holds

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