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    Hiya, so the Q is:
    f(x)=4-ln3x
    g(x)=e2-x
    show that fg(x)=x+a-lnb where a and b are integers to be found
    So fg(x)=4-ln3(e2-x)
    everything past this point is me just wildly guessing:
    4-ln(3e2/ex) -> 4-(ln3e2-lnex)
    4-ln3e2+x
    =x+4-ln3e^2
    This looks right up to here, then how do I simplify ln3e^2 to an integer?
    ln3e2=ln3+lne2=ln3+2...?
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    (Original post by Exephy)
    Hiya, so the Q is:
    f(x)=4-ln3x
    g(x)=e2-x
    show that fg(x)=x+a-lnb where a and b are integers to be found
    So fg(x)=4-ln3(e2-x)
    everything past this point is me just wildly guessing:
    4-ln(3e2/ex) -> 4-(ln3e2-lnex)
    4-ln3e2+x
    =x+4-ln3e^2
    This looks right up to here, then how do I simplify ln3e^2 to an integer?
    ln3e2=ln3+lne2=ln3+2...?
    Y = In x is the inverse function (log base e) of e^y , so if y = ln (e^x) then y = x as the function is basically undone by its inverse.
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    (Original post by kkboyk)
    Y = In x is the inverse function (log base e) of e^y , so if y = ln (e^x) then y = x as the function is basically undone by its inverse.
    I know, haha. But how does that help me here?
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    (Original post by Exephy)
    I know, haha. But how does that help me here?
    You basically apply that to any natural log function with an exponential function within it in your question. E.g. 3ln(e^(2-x)) = 3 (2-x)
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    (Original post by kkboyk)
    You basically apply that to any natural log function with an exponential function within it in your question. E.g. 3ln(e^(2-x)) = 3 (2-x)
    the problem is that the form the answer wants it in requires ln to be there :/
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    (Original post by Exephy)
    Hiya, so the Q is:
    f(x)=4-ln3x
    g(x)=e2-x
    show that fg(x)=x+a-lnb where a and b are integers to be found
    So fg(x)=4-ln3(e2-x)
    everything past this point is me just wildly guessing:
    4-ln(3e2/ex) -> 4-(ln3e2-lnex)
    4-ln3e2+x
    =x+4-ln3e^2
    This looks right up to here, then how do I simplify ln3e^2 to an integer?
    ln3e2=ln3+lne2=ln3+2...?
    You don't want ln(3e^2) to be an integer!

    Remember the question requires the final answer to have a logarithm in it. So all you need are 2 basic rules:

    ln(ab) = ln a + ln b
    ln(e^w) = w

    I haven't checked all your manipulations, but these 2 rules are all you need to get through the question
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    (Original post by davros)
    You don't want ln(3e^2) to be an integer!

    Remember the question requires the final answer to have a logarithm in it. So all you need are 2 basic rules:

    ln(ab) = ln a + ln b
    ln(e^w) = w

    I haven't checked all your manipulations, but these 2 rules are all you need to get through the question
    I got it, thanks!
 
 
 
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