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$\log_2(x) = \log_x(2) $

Using the change of base theorem: $\dfrac{\log(x)}{\log(2)} = \dfrac{\log(2)}{\log(x)}$

Multiplied the denominators on both sides: $\log(x)\log(x) = \log(2)\log(2)$

I kind of get stuck here. I know that you can't take the square root of both sides of the equation, but still, $x = 2$ seems to be an obvious solution to the equation. I've missed $2^{-1}$ or $\frac12$ as another answer to the equation, which I am struggling to get to.

Any help will be greatly appreciated, thanks in advance.

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    $\begingroup$ $(\log x)^2 = (\log 2)^2$ is equivalent to $(\log x - \log 2)(\log x + \log 2) = 0$, hence $\log x = \log 2$ or $\log x = - \log 2$, which give $x = 2$ and $x = \dfrac{1}{2}$ respectively. $\endgroup$ – Zhanxiong Jul 22 '15 at 3:09
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You have it, actually. \begin{align} (\log(x))^2 &= (\log(2))^2 \\ \log(x) &= \pm \log(2) \end{align}

For the "$+$" case, you've already solved it.

In the "$-$" case, you have $\log(x) = -\log(2) = \log(2^{-1}) = \log(\frac12)$, from which you can get $x=\frac12$.

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You manipulated the given equation and got the following equation: $[\log(x)]^2 = [\log(2)]^2$.

Move everything to the left side: $[\log(x)]^2 - [\log(2)]^2 = 0$.

Factor the left side: $[\log(x)-\log(2)][\log(x)+\log(2)] = 0$.

So, for the original equation to be true, you need either $\log(x)-\log(2) = 0$ or $\log(x)+\log(2) = 0$. Can you solve these two equations?

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HINT:

$$x^2=y^2\implies x=\pm y$$

and

$$-\log 2=\log (1/2)$$

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  • $\begingroup$ Why the down vote? $\endgroup$ – Mark Viola Jul 22 '15 at 15:02

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