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    Could someone please check what I'm doing for this logs question is right?
    I haven't done logs in ages!

    Thanks,
    Blake

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    For (a) be careful! The graph y = exp(3x) never touches the x-axis (y=0), it's an asymptote to the curve! So infact exp(3x) > 0 (strict inequality), and so the range is f(x) > 2(0) - 1 = -1

    (b) Yes what you have done is correct, but as we have exponentials (specifically e^x), we would usually use the natural logarithm, i.e. a logarithm base e (e = 2.718...). In this case log(e) = ln(e) = 1 as e^1 = e. Then as you had y = f(x), and rearranged for x = g(y), you can now see that f^{-1}(x) = g(x) which i what you have - just a different based logarithm
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    (Original post by Blake Jones)
    Could someone please check what I'm doing for this logs question is right?
    I haven't done logs in ages!

    Thanks,
    Blake

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    Very close for a - but when is f(x) equal to -1?

    Also better to write ln or log(base e) instead of just log - if you're using log base 10 then you won't get to the right answer.
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    (Original post by Blake Jones)
    Could someone please check what I'm doing for this logs question is right?
    I haven't done logs in ages!

    Thanks,
    Blake
    a. Nope. You cannot have it equal -1.

    b. Correct procedure, but just take the natural log instead of base 10 and the whole thing simplifies down.
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    (Original post by Chazatthekeys)
    For (a) be careful! The graph y = exp(3x) never touches the x-axis (y=0), it's an asymptote to the curve! So infact exp(3x) > 0 (strict inequality), and so the range is f(x) > 2(0) - 1 = -1

    (b) Yes what you have done is correct, but as we have exponentials (specifically e^x), we would usually use the natural logarithm, i.e. a logarithm base e (e = 2.718...). In this case log(e) = ln(e) = 1 as e^1 = e. Then as you had y = f(x), and rearranged for x = g(y), you can now see that f^{-1}(x) = g(x) which i what you have - just a different based logarithm
    Ah thank you! We haven't done about e or natural logs very much yet, only core 1 and core 2 stuff so I'm trying to remember haha!
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    (Original post by SeanFM)
    Very close for a - but when is f(x) equal to -1?

    Also better to write ln or log(base e) instead of just log - if you're using log base 10 then you won't get to the right answer.
    What do you do when differentiating logs? I've never done that before. (For part c )
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    (Original post by Blake Jones)
    What do you do when differentiating logs? I've never done that before. (For part c )
     \frac{d}{dx}ln(f(x)= \frac{f'(x)}{f(x)} eg for ln3x it is (3/3x) = 1/x.
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    (Original post by Blake Jones)
    What do you do when differentiating logs? I've never done that before. (For part c )
    Should you be doing these questions without having gone through the content then...?
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    (Original post by RDKGames)
    Should you be doing these questions without having gone through the content then...?
    I did question it but this is the homework we were given regardless xD
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    (Original post by SeanFM)
     \frac{d}{dx}ln(f(x)= \frac{f'(x)}{f(x)} eg for ln3x it is (3/3x) = 1/x.
    Right, okay I think I get that, I'll give it a go
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    (Original post by Blake Jones)
    I did question it but this is the homework we were given regardless xD
    Well if you want the proof to what good ol' Sean up there said, here it is:

    y=\ln[f(x)] \Rightarrow e^y=f(x)

    \Rightarrow \frac{d}{dx}e^y = \frac{d}{dx} f(x)

    \Rightarrow e^y \frac{dy}{dx} = f'(x)

    \frac{dy}{dx}=\frac{f'(x)}{e^y} \longrightarrow \frac{dy}{dx}=\frac{f'(x)}{f(x)}
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    (Original post by SeanFM)
     \frac{d}{dx}ln(f(x)= \frac{f'(x)}{f(x)} eg for ln3x it is (3/3x) = 1/x.
    Did you get the gradient to be 1?
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    (Original post by RDKGames)
    Well if you want the proof to what good ol' Sean up there said, here it is:

    y=\ln[f(x)] \Rightarrow e^y=f(x)

    \Rightarrow \frac{d}{dx}e^y = \frac{d}{dx} f(x)

    \Rightarrow e^y \frac{dy}{dx} = f'(x)

    \frac{dy}{dx}=\frac{f'(x)}{e^y} \longrightarrow \frac{dy}{dx}=\frac{f'(x)}{f(x)}
    Ah I see, thank you, that helps me understand it a bit better!
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    (Original post by Blake Jones)
    Did you get the gradient to be 1?
    I haven't calculated it myself, if you post a picture of your working I am happy to validate it
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    (Original post by SeanFM)
    I haven't calculated it myself, if you post a picture of your working I am happy to validate it
    Zero sorry! xD 1x0 is 0 Blake!

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    (Original post by Blake Jones)
    Zero sorry! xD 1x0 is 0 Blake!

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    1/3 is a constant, if you put it as part of f(x) then they would cancel.

    Derivative of (1/3) * lnx is (1/3) * the derivative of ln x for example.

    I'm not quite sure what you mean when you are using u , y etc to differentiate.
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    (Original post by SeanFM)
    1/3 is a constant, if you put it as part of f(x) then they would cancel.

    Derivative of (1/3) * lnx is (1/3) * the derivative of ln x for example.

    I'm not quite sure what you mean when you are using u , y etc to differentiate.
    I was trying to use the chain rule which I've just learned, could you show me what I should have done? I'm a bit confused.
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    (Original post by Blake Jones)
    I was trying to use the chain rule which I've just learned, could you show me what I should have done? I'm a bit confused.
    Derivative of (1/3 * f(x)) = (1/3) * f(x) so take that 1/3 out. Otherwise the 1/3 will appear in both f'(x) and f(x) thus cancelling out, when it shouldn't be.
    Otherwise you are just defining what the f(x) is when trying to differentiate ln(f(x)), then calculating (f'(x)/f(x)).

    I would say look up differentiating logs in your textbook before attempting this.
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    (Original post by SeanFM)
    Derivative of (1/3 * f(x)) = (1/3) * f(x) so take that 1/3 out. Otherwise the 1/3 will appear in both f'(x) and f(x) thus cancelling out, when it shouldn't be.
    Otherwise you are just defining what the f(x) is when trying to differentiate ln(f(x)), then calculating (f'(x)/f(x)).

    I would say look up differentiating logs in your textbook before attempting this.
    Will do, thank you very much!
 
 
 
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