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Solve for t - help please (probably pretty simple but I'm stuck) watch

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    I have two equations for the depreciation of 2 cars: -

    t being the number of years after after it is sold as new. The 15000 and 18000 are the initial values of the cars.

    V = 15000*e^-0.155*t
    W = 18000*e^0.175*t

    How can I find when V and W have depreciated to the same value?

    I started by putting 15000*e^-0.155*t = 18000*e^-0.175*t but now I'm stuck....

    Can any one help? Even if you just say breifly what I need to do that would prob help, I just need a nudge in the right direction I think.

    I presume it will have something to do with taking the natural log but I've tried various different ways and can't figure it out.
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    Looks like you started well. Put the numbers on one side and the exponentials on the other and ln everything.
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    Put the exponentials on the same side by using:

    \displaystyle \frac{e^a}{e^b} = e^{a-b}
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    (Original post by rosschambers1987)
    I have two equations for the depreciation of 2 cars: -

    t being the number of years after after it is sold as new. The 15000 and 18000 are the initial values of the cars.

    V = 15000*e^-0.155*t
    W = 18000*e^0.175*t

    How can I find when V and W have depreciated to the same value?

    I started by putting 15000*e^-0.155*t = 18000*e^-0.175*t but now I'm stuck....

    Can any one help? Even if you just say breifly what I need to do that would prob help, I just need a nudge in the right direction I think.

    I presume it will have something to do with taking the natural log but I've tried various different ways and can't figure it out.
    Divide both sides by one of the coefficients and both sides by the other e^-...value. getting e's on one side and a number on the other.
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    1=(6/5)(e^-0.175t-0.155t)
    5/6=e^(-0.33t)
    ln(5/6)=-0.33t
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    I think you can do it if you take the natural log of both equations to get

    

ln(w)-ln(18000)=0.175t

ln(v)-ln(15000)=-0.155t

    It says that w and v have depreciated to the same value so ln(v)=ln(w), ill let that equal C so you can see that it's just a simple simultaneous equation.

    

C-ln(18000)=0.175t

C-ln(15000)=-0.155t

    Now it's just a simple matter of solving the simultaneous equation. Hope that helps.

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ln(15000)-ln(18000)=-0.33t

ln(15000/18000)/-0.33=t

ln(5/6)/-0.33=t
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    (Original post by rosschambers1987)
    I have two equations for the depreciation of 2 cars: -

    t being the number of years after after it is sold as new. The 15000 and 18000 are the initial values of the cars.

    V = 15000*e^-0.155*t
    W = 18000*e^0.175*t

    How can I find when V and W have depreciated to the same value?

    I started by putting 15000*e^-0.155*t = 18000*e^-0.175*t but now I'm stuck....

    Can any one help? Even if you just say breifly what I need to do that would prob help, I just need a nudge in the right direction I think.


    I presume it will have something to do with taking the natural log but I've tried various different ways and can't figure it out.
    Firstly, do you mean W = 18000*e^-0.175? If so,

    let V = W.
    15000e^{-0.155t}=18000e^{-0.175t}

    cross dividing gives:
    \large \frac{18000}{15000}= \frac {e^{-0.175t}}{e^{-0.155t}}


    Now, looking at the right hand side. Using the rule: \frac {x^{a}}{x^{b}} = x^{(a-b)}
    RHS = e^{[-0.155t-(-0.175t)]} = e^{0.02t}
    LHS =  \frac {18000}{15000}= \frac {18}{15}= \frac {6}{5}

    Therefore, \frac {6}{5}=e^{0.02t}, logging both sides gives:-
    \large \ln \frac {6}{5} = \frac {t}{50}
    t=50\ln \frac {6}{5} or t=50\ln 6 - 50\ln5
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    Thanks to everybody for replying to my question. I now understand how to do it Very much appreciate your responses!
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    (Original post by reaped)
    RHS = e^[-0.155t-(-0.175t)] = e^0.02t
    -0.155-0.175 not-0.155-(-0.175)
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    (Original post by Vib-Rib)
    -0.155-0.175 not-0.155-(-0.175)
    I said at the start "Firstly, do you mean W = 18000*e^-0.175? If so,". Since he's talking about cars it would make sence that the price deteriorates over time.
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    (Original post by reaped)
    I said at the start "Firstly, do you mean W = 18000*e^-0.175? If so,". Since he's talking about cars it would make sence that the price deteriorates over time.
    ok sorry
 
 
 
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