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# Logs problem (p3 level) watch

1. see attached,

t= 22 but i cant show it

Thanks
Attached Images

2. Where did you get that question from? I can't solve it.

I get to t= (5 ^30)* e^(-1.5t)

how to solve that?
3. t is not equal to 22,plug it in to both sides and they will give different answers.
4. t=22 doesn't solve it
e^((ln22)/30) = 1.108..
5e^(-0.05*22) = 1.664..
Plotting it, it looks like t is somewhere around 30 (not exactly 30 though). I can't work out how to do it algebraically though.
5. (Original post by Bezza)
t=22 doesn't solve it
e^((ln22)/30) = 1.108..
5e^(-0.05*22) = 1.664..
Plotting it, it looks like t is somewhere around 30 (not exactly 30 though). I can't work out how to do it algebraically though.
i dont think its possible algebraically,i end up with
t^1/30 + 0.05t=ln5.
maybe u can use an iterative process?
6. The question is June01 AQA

Can be found here: http://www.aqa.org.uk/qual/gceasa/mat_assess.html

Q3 b)ii)

Q.

1000e^1/30ln2t = 5000e^-0.05t

thats the start of the question.

And the mark scheme reckons it goes to:

1/30ln2t + 0.05t = ln5 (but i cant get to this stage!!!)

Then they say t=22 for the 4th mark.
7. Thanks for your efforts guys.

I've managed to solve it !!

i forgot that when dividing powers you have to subtract powers!! e.g. e^-3/e^-2 = e^(-3)-(-2) = e^-1 BASIC school boy error!

Thats why i got it wrong!!

t=22.01!

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Updated: June 22, 2004
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