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solving an equation

  • 27-11-2008 7:56pm
    #1
    subzero12
    Closed Accounts Posts: 96 ✭✭


    10=8e^0.1t for t assuming t>0

    could anyone help me with this

    i got
    10=8e^-0.1t

    log 10=log8-0,1t lne
    0,1 t= -1+log8

    t= -10+10log8

    but i think im way off cause when you put it into the equation it doesnt equal 10


Comments

  • #2
    Cokehead Mother
    Closed Accounts Posts: 773 ✭✭✭Cokehead Mother


    You said 0.1t at the start but then changed to -0.1t so I dunno which one it's supposed to be.

    Assuming it's -0.1t, I think your mistake is saying log10 = 1 which is true if you mean log to the base 10 but then you're saying lne = 1. You should have taken the natural log of both sides.


  • #3
    subzero12
    Closed Accounts Posts: 96 ✭✭subzero12


    Cokehead Mother wrote: »
    You said 0.1t at the start but then changed to -0.1t so I dunno which one it's supposed to be.

    Assuming it's -0.1t, I think your mistake is saying log10 = 1 which is true if you mean log to the base 10 but then you're saying lne = 1. You should have taken the natural log of both sides.

    yeah it should be -0.1t at the start


  • #4
    subzero12
    Closed Accounts Posts: 96 ✭✭subzero12


    ln(10)=-0.1t*8 ln(e)
    ln(10)=-0.8t
    2.303=-0.8t
    t=-2.879

    i think i got it


  • #5
    LeixlipRed
    Closed Accounts Posts: 6,081 ✭✭✭LeixlipRed


    No not correct. Ok, are you sure what you're doing when you take natural logs on both sides? You've simplified the RHS wrongly after taking natural logs. Use the fact that (to any base) log(ab)=log(a) + log(b) and see what you get. Also, you should know you're wrong as if you sub you're answer back into the original equation you don't get 10 on the RHS.


  • #6
    Pherekydes
    Registered Users, Registered Users 2 Posts: 16,202 ✭✭✭✭Pherekydes


    LeixlipRed wrote: »
    No not correct. Ok, are you sure what you're doing when you take natural logs on both sides? You've simplified the RHS wrongly after taking natural logs. Use the fact that (to any base) log(ab)=log(a) + log(b) and see what you get. Also, you should know you're wrong as if you sub you're answer back into the original equation you don't get 10 on the RHS.

    Or even more obviously, assume t>0 (from OP). :D


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  • #7
    LeixlipRed
    Closed Accounts Posts: 6,081 ✭✭✭LeixlipRed


    Haha, didn't even think about that. So it has to be +0.1t in the original equation otherwise you end up with negative t. Some sort of equation detailing time travel possibly?? :D


  • #8
    nickcave
    Registered Users, Registered Users 2 Posts: 300 ✭✭nickcave


    Exp[-t/10] = 5/4

    -t/10 = Log[5/4]

    t = -10Log[5/4] = -2.23144


  • #9
    LeixlipRed
    Closed Accounts Posts: 6,081 ✭✭✭LeixlipRed


    Someone hasn't been paying attention. Just like me yesterday


  • #10
    nickcave
    Registered Users, Registered Users 2 Posts: 300 ✭✭nickcave


    yeah no solution in the positive quadrant I'm afraid. Plot of rhs:

    plots1vf7.jpg


  • #11
    LeixlipRed
    Closed Accounts Posts: 6,081 ✭✭✭LeixlipRed


    Well if you take t to mean time it would have to positve. Hence, the original equation must have a mis-print in it. It must be 10=8^e0.1t


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