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# Core 4: Exponential growth and decay help watch

1. If anyone can tell me how to do this question I would really appreciate it:

If two cars depreciate, one according to the model V=15000e^(-0.155t) and the other by W=18000e^(-0.175t), calculate the time,t, in years when the cars will have depreciated to the same value.

Thanks.
2. V=W
15000e^(-0.155t) = 18000e^(-0.175t)
5e^(-0.155t) = 6e^(-0.175t)
ln[5e^(-0.155t)] = ln[6e^(-0.175t)]
solve the equation ... if you need help, shout
3. (Original post by ludmila)
V=W
15000e^(-0.155t) = 18000e^(-0.175t)
5e^(-0.155t) = 6e^(-0.175t)
ln[5e^(-0.155t)] = ln[6e^(-0.175t)]
solve the equation ... if you need help, shout
Thanks, That's what I tried originally, but how do you solve it without the t's cancelling each other out?
4. (Original post by dan6780)
Thanks, That's what I tried originally, but how do you solve it without the t's cancelling each other out?
ln[5e^(-0.155t)] = ln[6e^(-0.175t)]
ln 5 + ln e
^(-0.155t) = ln 6 + ln e^(-0.175t)
ln 5 - 0.155t = ln 6 - 0.175t
0.02t = ln 6 - ln 5
0.02t = ln(6/5)
0.02t = ln 1.2
t = (ln1.2) / 0.02
t = 9.12 (to 2 decimal places)
5. (Original post by ludmila)
ln[5e^(-0.155t)] = ln[6e^(-0.175t)]
ln 5 + ln e^(-0.155t) = ln 6 + ln e^(-0.175t)
ln 5 - 0.155t = ln 6 - 0.175t
0.02t = ln 6 - ln 5
0.02t = ln(6/5)
0.02t = ln 1.2
t = (ln1.2) / 0.02
t = 9.12 (to 2 decimal places)
You are an absolute godsend! Thank you.
6. (Original post by dan6780)
You are an absolute godsend! Thank you.
You are more than welcome.

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