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    f(x) = log_2(3x + 1)

    Work out the inverse.

    Do i use the base changing formula for this? As i obviously need to get it to the base e, can someone show me how please.

    Cheers
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    Make it
    y = logV2(3x+1)
    2^y = (3x +1)
    (2^y) - 1 = 3x
    ((2^y) - 1) / 3 = x

    so the inverse is ((2^x) - 1) / 3

    That's what I'd do anyway. Hope it helps/is right!
    I have my Core 3 exam tomorrow, and I'm bricking it!!!
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    (Original post by dani_lou)
    Make it
    y = logV2(3x+1)
    2^y = (3x +1)
    (2^y) - 1 = 3x
    ((2^y) - 1) / 3 = x

    so the inverse is ((2^x) - 1) / 3

    That's what I'd do anyway. Hope it helps/is right!
    I have my Core 3 exam tomorrow, and I'm bricking it!!!
    It is right, yes.
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    (Original post by generalebriety)
    It is right, yes.
    Thank god for that!!!
    I may still have a chance tomorrow then!!!
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    (Original post by dani_lou)
    Make it
    y = logV2(3x+1)
    2^y = (3x +1)
    (2^y) - 1 = 3x
    ((2^y) - 1) / 3 = x

    so the inverse is ((2^x) - 1) / 3

    That's what I'd do anyway. Hope it helps/is right!
    I have my Core 3 exam tomorrow, and I'm bricking it!!!
    Oh right cheers dani, is there any rule/formula for this? Because i used to vaguely remember me doing this for C2, but wouldn't know why lol.
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    Well... The rule for finding the inverse, as far as I know, is tow make f(x) "y", and then re-arrange to make x the subject. Then you make all the ys xs, and that's your inverse.

    The log rule is that log-to-the-base-x (I can't be bothered to draw pictures!) is cancelled out by X^. If that makes any sense. You can only use e when you have ln (natural log).
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    (Original post by dani_lou)
    Well... The rule for finding the inverse, as far as I know, is tow make f(x) "y", and then re-arrange to make x the subject. Then you make all the ys xs, and that's your inverse.

    The log rule is that log-to-the-base-x (I can't be bothered to draw pictures!) is cancelled out by X^. If that makes any sense. You can only use e when you have ln (natural log).
    No, not that lol ... i meant getting the base of the logs and making it (2^y)
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    \log_a{b}=c



a^c=b
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    (Original post by Totally Tom)
    \log_a{b}=c



a^c=b
    Thanks Tom, picture drawing genius.
    To Doji... What he said! ^ ^ ^
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    (Original post by Totally Tom)
    \log_a{b}=c



a^c=b
    Oh thats it! cheers thanks tom
 
 
 
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