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Logs

"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.
Reply 1
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: log10z=log10y10kt\log_{10} z = \log_{10} y10^{-kt}

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

log10z=log10y+log1010kt\log_{10} z = \log_{10} y + \log_{10} 10^{-kt}

Now use the fact that logaam=mlogaa=m\log_a a^m = m \log_a a = m. Can you see how that gets you what you want?
(edited 8 years ago)
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: log10z=log10y10kt=log10y+log1010kt\log_{10} z = \log_{10} y10^{-kt}= \log_{10} y + \log_{10} 10^{-kt}

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

Now use the fact that logaam=mlogaa=m\log_a a^m = m \log_a a = m.


Thanks!
Reply 3
Original post by JustJusty
Thanks!


Got it all? Sure you understand everything?
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.
Reply 5
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. :biggrin:

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