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I want to know what values of x are so that $(-2)^x = 4$

If this equation is true, why $\log_{-2}{4}$ is undefined in $\mathbb{R}$?

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$\log_{-2}4=\frac{\ln4}{\ln(-2)}$, which leaves you with $\ln(-2)$, which is undefined in the real numbers.

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  • $\begingroup$ but if we use the definition that $(-2)^x=4$, I think that x would be 2. In that case, it's defined. $\endgroup$ – ABE_Mark45 Dec 3 '16 at 21:19
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Multiply by $-1$ the left term and take logarithms:

$\log_{2}(2^x) = \log_{2}(4) \Rightarrow x\log_{2}(2) = 2$

$x = 2$

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Consider the following:

Let $x=log_{-3}(27)$. Then, by definition $(-3)^x = 27$. This, however would lead to a contradiction, since $(-3)^3 = -27$. In that case, our function would not have a solution, which means it is not defined for the basis $-3$. Just because it works out for some special cases does not make it well defined.

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