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Differentiating Logarithmic and Exponential Functions

  1. Jun 1, 2006 #1
    a question i couldnt get the right answer to:

    1. y= 3^x (log(3) x)

    [3 = base (dont know the right syntax for typing up logs on the web)]

    what i have so far:
    y = 3^x (ln 3) (log(3) x) + 3^x / xln3

    right answer: 3^x (ln 3) + 3^x / xln3

    if we apply the product rule, where did the (log(3) x) go?
  2. jcsd
  3. Jun 1, 2006 #2


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    Are you sure the right answer isn't [tex](3^x)ln(x)+ \frac{3^x}{x\cdot ln(3)}[/tex]?

    In that case it's just a simplified form of your answer where you write [tex]log_3(x)[/tex] as [tex] \frac{ln(x)}{ln(3)}[/tex] and the ln(3) cancels out.
    Last edited: Jun 1, 2006
  4. Jun 1, 2006 #3
    yes, it is...you're absolutely right. thanks!
  5. Jun 2, 2006 #4
    quick question again, how does 2/ln4 = 1/ln2?
  6. Jun 2, 2006 #5


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    Well think about it. Can you use any of the properties of logs to write ln(4) in terms of ln(2)?
  7. Jun 2, 2006 #6
    ahhh...ln(4) = ln(2)^2 = 2ln(2).

    i have one more problem im having trouble differentiating

    y = x(3x)^(x^2) - i hope thats understandable (basically 3x to the power of x^2)

    so y' = (3x)^(x^2) + x(3x)^x^2*ln(3x)* 2x*3

    i just did product rule and chain rule but this is not even close to the right answer which is: x(3x)^x^2 * [1/x + x + 2x*ln(3x)]

    i dont know how they got 3 terms inside the square brackets when theres only one set of product rule (i.e f'g + fg')
  8. Jun 2, 2006 #7
    sorry one more:

    y = x^(1/ln(x))

    y' = 1/ln(x) * x^(1/ln(x) - 1)

    just doing power rule here but the answer is 0
  9. Jun 2, 2006 #8
    y = x^(1/ln(x))...
    Apply ln to both sides...
    ln y = 1 / ln x * ln x
    ln y = 1
    y'/y = 0 (Chain rule)
    y' = 0.
    Last edited: Jun 2, 2006
  10. Jun 2, 2006 #9


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    The fact that you find 0 means that y had to be a constant, right?
    Now if y = x^(1/ln(x)) then ln(y) = 1/ln(x).ln(x) = 1, as shown above.
    Well, if ln(y) = 1, then y has to be e so x^(1/ln(x)) is actually e :smile:
  11. Jun 2, 2006 #10


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    There's one more way, which can be applied to this problem. There's a property of logs faying:
    [tex]\log_a b = \frac{1}{\log_b a}[/tex]
    This can be proved by using the fact that 1 = logbb
    [tex]\Leftrightarrow \log_a b = \frac{\log_b b}{\log_b a}[/tex]
    [tex]\Leftrightarrow \log_a b = \log_a b[/tex] (which is true)
    So applying that property here, we have:
    [tex]x ^ {\frac{1}{\ln x}} = x ^ {\log_x e} = e[/tex]
    Now, can you differentiate it? :)
  12. Jun 2, 2006 #11
    My way's the coolest. ;)
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