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IsTim

  • 2 years ago

Solve 3^x=15 by taking the natural logarithm of both sides.

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  1. LordHades
    • 2 years ago
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    x log3 = log15 x = log15/log3

  2. roadjester
    • 2 years ago
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    The natural logarithm is ln not log

  3. roadjester
    • 2 years ago
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    \[3^x = 15\] \[\ln 3^{3^x} =\ln 3^{15}\] \[x=15\ln3\]

  4. roadjester
    • 2 years ago
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    Is that the answer you have?

  5. IsTim
    • 2 years ago
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    I just didn't know what to to do. Thanks though, Roadjester.

  6. roadjester
    • 2 years ago
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    Do you know the Laws of Logs?

  7. roadjester
    • 2 years ago
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    They are the same for natural log, just that you use natural log for the change of base formula.

  8. IsTim
    • 2 years ago
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    What's the difference between natural and common logarithms? I think people may have applied the incorrect technique into my other question.

  9. roadjester
    • 2 years ago
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    Well, this is the general form for log: \[\log_{a}x \] where x is a variable and a is the base. This is basically the inverse of an exponential function of the form:\[x = a^y\]. You'll notice that rather than writing the usual "y is a function of x" I have it reversed. If you're familiar with inverse functions, this shouldn't be too troublesome.

  10. roadjester
    • 2 years ago
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    Now as for what the difference between \[\ln x\] and \[\log_{} x\] is, you'll notice that when I wrote log x, I had a base a. log can take on any base. The natural log is assumed to have the base of e which is approximately 2.7182818...

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