## 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)

1. anonymous

2. anonymous

divide by e^(0.05t) on both sides

3. anonymous

and post what you get.

4. anonymous

ok hold on

5. anonymous

17tln(0.07)-tln(0.05)=11

6. anonymous

$e ^{x} / e ^{y} = e ^{x-y}$$\ln{e^{x-y}} = x - y$

7. anonymous

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

I did this : 17e^(0.07t)/(e^(0.05t) = 11

9. anonymous

ok, thar part is right. what did you do after that?

10. anonymous

okay, and then?

11. anonymous

is it possible that you can take a picture of your notes and attach it here? that way it is much faster.

12. anonymous

according to the quotient rule u subtract?

13. anonymous

right, you subtract

14. anonymous

so what did you get?

15. anonymous

17tln(0.07) - tln(0.05)=11

16. anonymous

why did you do tln0.07-tlon0.05? what rule is that?

17. anonymous

idk. can u show me the next step. plz

18. anonymous

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

hold on let me work it out

20. anonymous

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

21. anonymous

i understand the simplification

22. anonymous

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

the rule is e^x/e^y = e^(x-y), not e^x/y = e^(x-y)

24. anonymous

yep...caught myself there :)

25. anonymous

yup. qoptient rules for natural logs right?

26. anonymous

lol. no its the quotient rule for ALL fractions.

27. anonymous

yep....now if you'll excuse me, i have to go make some natural logs of my own....;o

28. anonymous

TMI

29. anonymous

hahaha

30. anonymous

suprisingly im an engineering major

31. anonymous

lol

32. anonymous

surprisingly, i earned a bachelor's in biochemistry....when i wasnt drunk...