Basic Logarithm Equation

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 $x=\frac12$.


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?


HINT:

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

and

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

Tags:

Logarithms

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