- anonymous

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

- katieb

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- anonymous

HELP ME PLEASE

- anonymous

divide by e^(0.05t) on both sides

- anonymous

and post what you get.

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## More answers

- anonymous

ok hold on

- anonymous

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

- anonymous

\[e ^{x} / e ^{y} = e ^{x-y}\]\[\ln{e^{x-y}} = x - y\]

- 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.

- anonymous

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

- anonymous

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

- anonymous

okay, and then?

- anonymous

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

- anonymous

according to the quotient rule u subtract?

- anonymous

right, you subtract

- anonymous

so what did you get?

- anonymous

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

- anonymous

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

- anonymous

idk. can u show me the next step. plz

- 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

- anonymous

hold on let me work it out

- anonymous

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

- anonymous

i understand the simplification

- 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

- anonymous

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

- anonymous

yep...caught myself there :)

- anonymous

yup. qoptient rules for natural logs right?

- anonymous

lol. no its the quotient rule for ALL fractions.

- anonymous

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

- anonymous

TMI

- anonymous

hahaha

- anonymous

suprisingly im an engineering major

- anonymous

lol

- anonymous

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

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