What are the values of x?

Question

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 $\mathbb{R}$?

Check this out:http://math.stackexchange.com/questions/690024/why-must-the-base-of-a-logarithm-be-a-positive-real-number-not-equal-to-1 — AxiomaticApproach, Dec 03 '16 at 21:02

score 0 · Answer 1 · edited Dec 03 '16 at 21:01

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Multiply by $-1$ the left term and take logarithms:

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

$x = 2$

edited Dec 03 '16 at 21:01

AlgorithmsX

4,560

answered Dec 03 '16 at 20:58

kub0x

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score 0 · Answer 2 · answered Dec 03 '16 at 20:59

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

answered Dec 03 '16 at 20:59

AlgorithmsX

4,560

but if we use the definition that $(-2)^x=4$, I think that x would be 2. In that case, it's defined. – ABE_Mark45 Dec 03 '16 at 21:19

score 0 · Answer 3 · answered Dec 03 '16 at 21:08

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.

What are the values of x?

3 Answers3