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    "z=y10^(-kt), where y and k are positive constants.

    i) Give the physical interpretation for the constant y."
    I have no idea what this means but I wrote ylog_10_z = -kt

    "ii) Show that log_10_z = -kt+log_10_y"
    By my workings, it really doesn't, so something is wrong. I've been stuck on this for days and it's due tomorrow.

    Note: I know all the logarithm rules, but none of them seem to apply here.
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    (Original post by JustJusty)
    "z=y10^(-kt), where y and k are positive constants.

    i) Give the physical interpretation for the constant y."
    I have no idea what this means but I wrote ylog_10_z = -kt
    What is z when t=0? Does that give you any clues as to the meaning of y?

    Hint: initial...

    "ii) Show that log_10_z = -kt+log_10_y"
    By my workings, it really doesn't, so something is wrong. I've been stuck on this for days and it's due tomorrow.

    Note: I know all the logarithm rules, but none of them seem to apply here.
    Take the base-ten logarithm of both sides: \log_{10} z = \log_{10} y10^{-kt}

    Use the fact that \log_a bc = \log_a b +\log_a c. (i.e: addition rule):

    \log_{10} z = \log_{10} y + \log_{10} 10^{-kt}

    Now use the fact that \log_a a^m = m \log_a a = m. Can you see how that gets you what you want?
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    (Original post by Zacken)
    What is z when t=0? Does that give you any clues as to the meaning of y?

    Hint: initial...



    Take the base-ten logarithm of both sides: \log_{10} z = \log_{10} y10^{-kt}= \log_{10} y + \log_{10} 10^{-kt}

    NB: For the last step, we used the fact that \log_a bc = \log_a b +\log_a c. (i.e: addition rule).

    Now use the fact that \log_a a^m = m \log_a a = m.
    Thanks!
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    (Original post by JustJusty)
    Thanks!
    Got it all? Sure you understand everything?
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    (Original post by Zacken)
    Got it all? Sure you understand everything?
    Yup, I've written it all out and got the right answer in the end. I guess I just couldn't figure out how to start it, and I kept trying to use my answer to part (i).

    You've been really helpful, thank you.
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    (Original post by JustJusty)
    Yup, I've written it all out and got the right answer in the end. I guess I just couldn't figure out how to start it, and I kept trying to use my answer to part (i).

    You've been really helpful, thank you.
    That's great then, good job! :woo:

    Thank you.
 
 
 
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