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    Find inverse function of: f(x) = e^(x-2) - 1

    So I did:

    f(x) + 1 = e^(x-2)
    f(x) + 1 = e^x multiplied by e^-2
    lnf(x) + ln1 = -2x

    gives x=-lnf(x)/2

    Then swap f(x) for x and make x become f-1

    f-1(x) = ln(x^-1/2)

    Is this correct?

    Thanks again!
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    I would've done this?
    y=e^(x-2) - 1

    y+1 = e^(x-2)
    ln (y+1) = (x-2) ln e
    x= 2 + ln ( y+1)

    therefore f-(x) = 2+ ln (x+1)

    ( we're taught to rearrange then swap x and y and the end)

    not sure about it though!
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    No, we require a function f such that x = e^(f(x)-2) - 1.

    Rearranging gives f(x) = ln(x+1) +2.
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    y = e^(x - 2) - 1

    Swap x and y:

    x = e^(y - 2) - 1

    Move 1 over:

    x + 1 = e^(y - 2)

    Taking logs of both sides:

    ln (x + 1) = y - 2

    Move 2 over:

    y = 2 + ln (x + 1)
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    Yeah, agree with all the above replies
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    (Original post by Chinesegirl)
    Yeah, agree with all the above replies
    Ok thanks everyone - think I need some more practise!
 
 
 
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