$-49 =7^x$ is the question. Here I m supposed to solve for what power of $7$ will give me $-49$. Or in other words, I have to solve for $x$. This looks fairly simply when thinking about the exponent rules for it looks as if you could make the $-49$ become $7$ to the power of $2$ so that the bases are common. However, it is a negative so I can't do this. I am unsure how to approach this simply because of the negative on the left side of the expression. Please help me out.
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1$a^x>0$ when $a>0$ – David P Sep 29 '14 at 17:19
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Sounds like complex: http://www.wolframalpha.com/input/?i=7%5Ex%3D-49&dataset= – MattAllegro Sep 29 '14 at 17:21
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is $x$ a real number? – Dr. Sonnhard Graubner Sep 29 '14 at 17:47
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This can not be a real number. – Michael Ejercito Feb 07 '24 at 17:06
1 Answers
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$7^x = e^{x \ln 7}$.
Now you want $$e^{x \ln 7} = -49 = e^{\Re (x \ln 7)} e^{i \Im (x \ln 7)} \equiv e^R e^{i \theta}$$
where $R \equiv \Re(x) \ln 7$ and $\theta \equiv \Im(x) \ln 7$.
Since the final form of the product can be expressed as the product of a magnitude and direction in the complex plane, and we know the magnitude is 49 and direction is left, we have $$ \begin{align} e^R &= 49 \implies \Re(x) = \frac{\ln 49}{\ln 7} = 2 \\ e^{i \theta} &= -1 \implies \Im(x) = \frac{\pi + 2 \pi n}{\ln 7} \end{align} $$ where $n$ is any integer.
Thus $x = 2 + i \frac{2 \pi (n+1/2)}{\ln 7}$ for any integer $n$.
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