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Exponential Growth for Pre-Calculus

  1. May 24, 2009 #1
    Hi everyone.

    My final is coming up for Precalc, and I'm studying my butt off.

    I was really needing help with this Exponential Growth Equation to find variable t.

    8.0e^(.033t) = 59.8e^(.001t)

    (8 times e to the .033t equals 59.8 times e to the .001t)

    I would greatly appreciate this because I'm stressing out for my finals. I have always been bad at this stuff
     
  2. jcsd
  3. May 24, 2009 #2

    CRGreathouse

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    Just take the logarithm of both sides and it's easy to solve.
     
  4. May 24, 2009 #3
    I know it sounds like I'm asking you to do it for me, but I really don't know where to begin. I'm confuzed because of the coefficients, because i can just natural log both sides if there weren't any.

    If you can please guide me good sir, I KNOW I can do the rest of this subchapter by myself. Thanks a bunch
     
  5. May 24, 2009 #4
    I know it's simple for you, but I swear it's so complicated for me. Please, this would really help me my kind sir
     
  6. May 24, 2009 #5

    Integral

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    What do you know about logs?

    Review the basic laws, look especially at logs and multiplication. Also study the relationship between e and ln.
     
  7. May 24, 2009 #6
    I know about logs my good sir. It's just this one problem tripping me up. I just really need someone to show me how its done, and your help will defaintely be appreciated. Should i make one side equal zero? i just dont know,
     
  8. May 24, 2009 #7
    will someone please help, this is the last problem i need and im done with this sub chater
     
  9. May 24, 2009 #8
    Please help me with this 1 equation. Then i'll be done with this subchapter

    Please, I really need help with this.

    I have to solve for t here.

    8.0e^( .033t ) = 59.8e^ (.001t)

    Please, i just really need you to show me how this is done.
     
  10. May 24, 2009 #9

    Borek

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    Re: Please help me with this 1 equation. Then i'll be done with this subchapter

    exey = ex+y
     
  11. May 24, 2009 #10
    Re: Please help me with this 1 equation. Then i'll be done with this subchapter

    I know that, but this is a weird one with weird coefficients. Please just walk me through this. gosh, im stressing out already
     
  12. May 24, 2009 #11
    So CRGreathouse is suggesting to do this:

    [tex]8.0e^{.033t} = 59.8e^{.001t}[/tex]

    [tex]\ln{(8.0e^{.033t})} = \ln{(59.8e^{.001t})}[/tex]

    Tell us what to do next.


    01
     
  13. May 24, 2009 #12
    so will it then turn into

    ln 8.0 (.033t) = ln 59.8 (.001t)

    Please ive been wait all day just to solve this one dang problem.
     
  14. May 24, 2009 #13

    Integral

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    No, You need to use the laws for logs and multiplication.
     
  15. May 24, 2009 #14
    integral, i beg of you. i just really need someone to help me with this mroe in depth. i've been waiting all day, i would really appreciate it integral. please
     
  16. May 24, 2009 #15
    would someone pklease help me. Im begign you guys. I need help with this. geeze
     
  17. May 24, 2009 #16
    Re: Please help me with this 1 equation. Then i'll be done with this subchapter

    would someone please help me im freaking begging you. im down on my knees
     
  18. May 24, 2009 #17
    WOULD SOMEONE PLEASE HELP ME

    im begging yiou guys
     
  19. May 24, 2009 #18

    Integral

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    We need to see some effort on your part. Show me that you have even tried to apply the hints you have been given.

    Your problem is of the form:

    [tex] A e^x = B e^y [/tex]
    So taking the ln:
    [tex] ln(A e^x) = ln (B e^y) [/tex]

    since ln(a * b ) = lna + lnb
    [tex] lnA + ln(e^x) = lnB + ln(e^y) [/tex]

    Can you finish?
     
  20. May 24, 2009 #19
    Integral, thank you kind sir.

    Let's see here.

    Yes sir.

    Looks like it will be

    ln 80 + .033t = ln 59.8 + .001t

    From there

    ln 80 - ln 59.8 = .001t - .033t

    ln 80 - ln 59.8 = -.032
    then divide and use a calculatro


    something isnt adding up quite right sir, im getting a negattive numbe[/I]

    wait, edit, it's ln 8 not 80. Thank you integral YOU ARE SO NICE. thank u so much.
     
    Last edited: May 24, 2009
  21. May 24, 2009 #20

    HallsofIvy

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    You have been helped repeatedly. How about doing something yourself?\
    You have been told exactly what to do. Why do you keep whining "do it for me! Do it for me!"?
     
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