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anonymous

  • 5 years ago

solve for t: 17e^(0.07t)=11e^(0.05t) I have so far gotten to this step: 17t ln(0.07)=tln(0.05)

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  1. anonymous
    • 5 years ago
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    HELP ME PLEASE

  2. anonymous
    • 5 years ago
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    divide by e^(0.05t) on both sides

  3. anonymous
    • 5 years ago
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    and post what you get.

  4. anonymous
    • 5 years ago
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    ok hold on

  5. anonymous
    • 5 years ago
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    17tln(0.07)-tln(0.05)=11

  6. anonymous
    • 5 years ago
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    \[e ^{x} / e ^{y} = e ^{x-y}\]\[\ln{e^{x-y}} = x - y\]

  7. anonymous
    • 5 years ago
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    how did you get to the step that you got to? what did you do to get 17t ln(0.07)=tln(0.05)? because that is incorrect.

  8. anonymous
    • 5 years ago
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    I did this : 17e^(0.07t)/(e^(0.05t) = 11

  9. anonymous
    • 5 years ago
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    ok, thar part is right. what did you do after that?

  10. anonymous
    • 5 years ago
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    okay, and then?

  11. anonymous
    • 5 years ago
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    is it possible that you can take a picture of your notes and attach it here? that way it is much faster.

  12. anonymous
    • 5 years ago
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    according to the quotient rule u subtract?

  13. anonymous
    • 5 years ago
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    right, you subtract

  14. anonymous
    • 5 years ago
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    so what did you get?

  15. anonymous
    • 5 years ago
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    17tln(0.07) - tln(0.05)=11

  16. anonymous
    • 5 years ago
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    why did you do tln0.07-tlon0.05? what rule is that?

  17. anonymous
    • 5 years ago
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    idk. can u show me the next step. plz

  18. anonymous
    • 5 years ago
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    alright so 17e^0.07t = 11e^0.05t Step 1: divide both sides by e^0.05t ...so you have (17e^0.07t)/(e^0.05t) = 11 Step 2: since e^x/y = e^x-y ...you can make this 17e^(0.07t-0.05t) =11 ...which equals???? 17e^0.02t = 11 Step 3: divide 17 out of there ...so e^0.02t = (11/17) step 4: take the natural log of both sides ...so ln e^0.02t = ln (11/17) step 5: e goes away now and you have 0.02t = ln (11/17) step 6: divide both sides by 0.02 so now it is t = (ln (11/17)/(0.02)) SO NOW..... t = -21.76590356 Check to see if this works in the original equation

  19. anonymous
    • 5 years ago
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    hold on let me work it out

  20. anonymous
    • 5 years ago
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    yup it works. i understand now, i was not certain at first if u can use the e^x/y rule

  21. anonymous
    • 5 years ago
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    i understand the simplification

  22. anonymous
    • 5 years ago
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    i messed up in my steps by going too fast and said e^x/y = e^x-y i meant to say (e^x) / (e^y) = e^x-y but i think you already figured that out

  23. anonymous
    • 5 years ago
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    the rule is e^x/e^y = e^(x-y), not e^x/y = e^(x-y)

  24. anonymous
    • 5 years ago
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    yep...caught myself there :)

  25. anonymous
    • 5 years ago
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    yup. qoptient rules for natural logs right?

  26. anonymous
    • 5 years ago
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    lol. no its the quotient rule for ALL fractions.

  27. anonymous
    • 5 years ago
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    yep....now if you'll excuse me, i have to go make some natural logs of my own....;o

  28. anonymous
    • 5 years ago
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    TMI

  29. anonymous
    • 5 years ago
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    hahaha

  30. anonymous
    • 5 years ago
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    suprisingly im an engineering major

  31. anonymous
    • 5 years ago
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    lol

  32. anonymous
    • 5 years ago
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    surprisingly, i earned a bachelor's in biochemistry....when i wasnt drunk...

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