Why does this equation hold

**bakerconspiracy** · June 10th 2012, 10:13 PM

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

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

thanks for the help in advance.

**Prove It** · June 10th 2012, 10:17 PM

Originally Posted by bakerconspiracy

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

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

thanks for the help in advance.

It only holds true for the values of n such that n^2 = 2^n.

**bakerconspiracy** · June 10th 2012, 10:31 PM

In my homework we have a solution that reduces from $2^{lg^2(n) } = 2^{lg(n) * lg(n)} = n^{lg(n)}$ where $lg(n) = log_2(n)$ . 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.

**Prove It** · June 10th 2012, 10:51 PM

Originally Posted by bakerconspiracy

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

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

thanks for the help in advance.

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

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

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

Q.E.D.

**Soroban** · June 11th 2012, 06:06 AM

Hello, bakerconspiracy!

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

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

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

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

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

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

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