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MDI Jade 7.3 XRD Software and Learning. MDI is best known for JADE and is used in labs wherever X-ray powder diffraction data are explored.Q:

Find x such that $\log_2{x} = 1+i\sqrt{2}$

Question: Let $x$ be a real number. Find all $x$ such that $\log_2{x} = 1+i\sqrt{2}$

What I have done:

Let $z=\log_2x$
$x=e^{2z}$
$x=2^z$
$z=\log_2x=\log_2(2^z)=\log_22^z$
$z=\log_22+i\sqrt{2}\log_22$
$z=\log_2(2)+i\sqrt{2}\log_2(2)$
$z=\frac{1}{\sqrt{2}}+i\sqrt{2}\frac{1}{\sqrt{2}}$
$z=\frac{1}{\sqrt{2}}(1+i\sqrt{2})$
$z=\frac{1}{\sqrt{2}}(1+i\sqrt{2})=\frac{1}{\sqrt{2}}(1+i(2+\sqrt{2}))=\frac{1+i}{\sqrt{2}}(1+2i)$
$z=\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}(1+2i)$

I’m not sure if this is the correct way to go about it?
I’m just doing it by plugging values in, as the book states, but some of these were difficult to figure out and I’m not sure if I should of found the answer before plugging it in the questions.

A:

We have
$$z=\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}(1+2i)\implies z^2=1+4i$$
which has two solutions, which are
2i,\, 1+i\ https://epkrd.com/macos-ux-pack-crack-lifetime-activation-code-download-for-windows-2022-latest/

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