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Math Help - Derivative of an exponential logarithm?

  1. #1
    Junior Member NAPA55's Avatar
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    Derivative of an exponential logarithm?

    New topic we started today, and like always the lesson is easy and straightforward but the homework is more complicated. The examples we did in class weren't like this question.

    We learned how to find the derivative of an exponential function, base b. It is the original function multiplied by ln b multiplied by the derivative of the exponent.

    We also learned how to find the derivative of a logarithmic function. It is the derivative of the function itself devided by the original function multiplied by ln b.

    Ok, that's all well and good, but what about when you have to find the derivate of the logarithmic function that has an exponential function in place of x.

    Find the derivative of this function:
    f(x) = 2log3(5^x) - log3(4^x)

    Where 3 is the base of the logarithmic function (I don't know how to make it a subscript).

    Thanks for any help. This is the only type of question I didn't get from today's homework.
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  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    f(x) = 2log3(5^x) - log3(4^x)
    using log properties

    2log_3(5^x)=2xlog_3(5)=(2log_3(5)) \cdot x

    All the stuff in front of the x is constant so

    \frac{df}{dx}=2log_3(5)
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  3. #3
    Junior Member NAPA55's Avatar
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    Thanks, I'll run that out and see if I get it right.
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  4. #4
    Junior Member NAPA55's Avatar
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    Not getting it right for some reason. This is what I did:

    <br />
p = 2log_3(5^x) - log_3(4^x)

     p' = 2(log_3(5)) - (log_3(4))

     p' = 2 (\frac{1}{5ln3}) -  (\frac{1}{4ln3})

     p' = \frac{2}{5ln3} - \frac{1}{4ln3}

    The answer should be

     p' = \frac {2ln5 - ln4}{ln3}
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by NAPA55 View Post
    Not getting it right for some reason. This is what I did:

    <br />
p = 2log_3(5^x) - log_3(4^x)

     p' = 2(log_3(5)) - (log_3(4))

     p' = 2 (\frac{1}{5ln3}) -  (\frac{1}{4ln3})
    How did you get that third line? That's where your problem is. You are applying the change of base formula to change the base from 3 to e:
    log_3(5) = \frac{ln(5)}{ln(3)}

    Try this in the other term and you'll get the correct answer.

    -Dan
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  6. #6
    Junior Member NAPA55's Avatar
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    I eventually realized that.

    It's all good now.
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