logx^1/64 =-3/2

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logx^1/64 =-3/2

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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Assuming I interpret your notation correctly,\[\log{x^{\frac{1}{64}}}=-\frac{3}{2}\rightarrow \frac{1}{64}\log{x}=-\frac{3}{2}\]So\[\log{x}=64 \times -\frac{3}{2}=-96\]Hence,\[x=e^{-96}\]
This is log base 10 not log base e or natural log so the answer should be: \[x=10^{-96}\]
Doesn't have to be.

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Yeah it does unless you change the base for x. log and ln are not equivalent. put it in a TI-89 or some math program
Well, it's a matter of definition. Log is technically the notation for natural logarithm in general, with ln reserved for use as the natural logarithm of the magnitude of a complex number. It's semantics. Mathematicians use log.
in maths log means base 10 not e unless it is mentioned.
Whatever. I have a degree + postgrad. in the subject. I'm not arguing about it anymore. Mathematicians use log for ln unless the base is made explicit.
srry then.

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