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    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.

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    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
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    (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?
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    (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)
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    (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.
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    (Original post by dan6780)
    You are an absolute godsend! Thank you.
    You are more than welcome.
 
 
 
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