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anonymous

  • 5 years ago

logx^1/64 =-3/2

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  1. anonymous
    • 5 years ago
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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,\[x=e^{-96}\]

  2. anonymous
    • 5 years ago
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    This is log base 10 not log base e or natural log so the answer should be: \[x=10^{-96}\]

  3. anonymous
    • 5 years ago
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    Doesn't have to be.

  4. anonymous
    • 5 years ago
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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

  5. anonymous
    • 5 years ago
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    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.

  6. anonymous
    • 5 years ago
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    in maths log means base 10 not e unless it is mentioned.

  7. anonymous
    • 5 years ago
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    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.

  8. anonymous
    • 5 years ago
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    srry then.

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