Exponential revision question I can't work out?

[Ramjam]

< Mentor Note -- Thread moved from the Calculus forum to HH/Calculus >[/color]

Hey everyone, I've started my new course and have been given revision and tutorial questions to do at home, this question has been bugging me for days now! I need to make "t" the subject to allow me to progress into the solution further.

Make t the subject in " i=12.5(1-e^-t/cr) "
I know that i need to use natural logs (ln) however i cannot seem to transpose the equation correctly. Could you please help me transpose this.

Thanks
Ramjam

p.s forgot to add my working so far, DUH!
ive transposed it to this form so far
i/12.5 = 1 - e ^-t/cr

this is where the logs come in and i have trouble, I am not sure what to do next?!?

 

[pasmith]

Before taking logs, you must make e^{-t/cr} the subject.

 

[Mark44]

< Mentor Note -- Thread moved from the Calculus forum to HH/Calculus >

Hey everyone, I've started my new course and have been given revision and tutorial questions to do at home, this question has been bugging me for days now! I need to make "t" the subject to allow me to progress into the solution further.

Make t the subject in " i=12.5(1-e^-t/cr) "

This is very unclear.
The quantity in parentheses could be any of the following:
$1 - \frac{e^{-t}}{cr}$
$1 - \frac{e^{-t}}{c}r$
$1 - e^{-t/c * r}$

I know that i need to use natural logs (ln) however i cannot seem to transpose the equation correctly. Could you please help me transpose this.

Thanks
Ramjam

p.s forgot to add my working so far, DUH!
ive transposed it to this form so far
i/12.5 = 1 - e ^-t/cr

this is where the logs come in and i have trouble, I am not sure what to do next?!?

No, logs don't come in yet. You still have that 1 on the right side.

 

[Ramjam]

How do I go about removing the one on the right hand side, is it correct for me to move it to the left hand side, leaving e^-t/cr on the right?

 

[Ray Vickson]

This is very unclear.
The quantity in parentheses could be any of the following:
$1 - \frac{e^{-t}}{cr}$
$1 - \frac{e^{-t}}{c}r$
$1 - e^{-t/c * r}$
No, logs don't come in yet. You still have that 1 on the right side.

He might even mean $\frac{1-e^{-t}}{cr}$. For some unknown reason, he refuses to deal with your question.

 

[Ramjam]

My lecturer hinted that the 1 should be moved leaving e^t/cr on the right, however unsure how to transpose it to that

 

[vela]

< Mentor Note -- Thread moved from the Calculus forum to HH/Calculus >

I moved this to the pre-calc forum since the issue is apparently about basic algebra.

 

[Mark44]

My lecturer hinted that the 1 should be moved leaving e^t/cr on the right, however unsure how to transpose it to that

Please address the question I asked in post #3.

 

[Ramjam]

Please address the question I asked in post #3.

1-e^-t/cr sorry my phone cut your post so I didn't understand what you meant

 

[Mark44]

You have

i/12.5 = 1 - e ^-t/cr

What can you do to both sides of this equation so that the 1 on the right side is gone? Possible choices are to add the same number toboth sides, subtract the same number from both sides, multiply both sides by the same nonzero number, or divide both sides by the same nonzero number.

 

[Ramjam]

If I was to add 1 to both sides e.g i/12.5 + 1 = 1-1e^-t/cr is this acceptable? As then the one can be removed from the right ?

 

[Mark44]

If I was to add 1 to both sides e.g i/12.5 + 1 = 1-1e^-t/cr is this acceptable?

No, plus it's wrong.
i/12.5 = 1 - e^-t/cr

If you add 1 to both sides, you get
i/12.5 + 1 = 1 - e^-t/cr + 1 = 2 - e^-t/cr
So that's no help at all.

Note that 1e^-t/cr is exactly the same as e^-t/cr, so there's no point in writing a coefficient of 1 here.

As then the one can be removed from the right ?

Nope.

 

[Ramjam]

Could you possibly take me through step by step how to remove 1 from the right leaving e-t/cr so that I can see how it's done, as confusing my self now

Ramjam

 

[Mark44]

Could you possibly take me through step by step how to remove 1 from the right leaving e-t/cr so that I can see how it's done, as confusing my self now

There is no "step by step" here - this is not a complicated process. Earlier in this thread you said,

If I was to add 1 to both sides e.g i/12.5 + 1 = 1-1e-t/cr is this acceptable?

You added 1 to the left side, but you didn't add 1 to the right side. Whatever you do to one side of the equation, you have to do the same to the other side.

If you have i/12.5 = 1 - e-^t/cr, what can you add to the 1 on the right side to get rid of it? In other words, what + 1 = 0?

 

[Ramjam]

There is no "step by step" here - this is not a complicated process. Earlier in this thread you said,
You added 1 to the left side, but you didn't add 1 to the right side. Whatever you do to one side of the equation, you have to do the same to the other side.

If you have i/12.5 = 1 - e-^t/cr, what can you add to the 1 on the right side to get rid of it? In other words, what + 1 = 0?

Okay so adding -1 to both sides i/12.5 -1 = 1-e^-t/cr -1
which when becomes i/12.5 -1 = -e^-t/cr

isnt the minus infront the the "e" become an issue? as you can log "-e"

Thankyou for your help on the matter

 

[mfb]

You can still multiply both sides by the same factor...
(and you should try those basic steps yourself)

 

[Mark44]

Okay so adding -1 to both sides i/12.5 -1 = 1-e^-t/cr -1
which when becomes i/12.5 -1 = -e^-t/cr

Yes.

isnt the minus infront the the "e" become an issue? as you can log "-e"

So what can you do to get rid of the minus sign? Remember to do the same thing to both sides of the equation.

 

[Ramjam]

Yes.
So what can you do to get rid of the minus sign? Remember to do the same thing to both sides of the equation.

Umm... I am not actually to sure? :/

 

[mfb]

Umm... I am not actually to sure? :/

Then try different things. At least one of them will work.

 

[Ramjam]

Then try different things. At least one of them will work.

Well I've been trying things for the past half hour with no luck, I am not seeing how i can multiply both sides to remove the minus from the "e"

 

[mfb]

Well I've been trying things for the past half hour with no luck

Then show your work please.

What could you multiply a negative number with to get a positive number?

 

[Ramjam]

Then show your work please.

What could you multiply a negative number with to get a positive number?

By multiplying -1 to let's say -1 you get 1, therefore I can see that by multiplying -1 to my equation will result in the right hand side becoming
e^-t/cr

However if i was then to multiply the other side as required i don't see what i will get?

 

[Mark44]

By multiplying -1 to let's say -1 you get 1, therefore I can see that by multiplying -1 to my equation will result in the right hand side becoming
e^-t/cr

However if i was then to multiply the other side as required i don't see what i will get?

Show us what you get when you multiply the left side by -1.

 

[Ramjam]

Show us what you get when you multiply the left side by -1.

- l/12.5 -1 ?

 

[Mark44]

- l/12.5 -1 ?

OK, what's the complete equation?

 

[Ramjam]

- l/12.5-1=e^-t/cr

However doesn't this cause another issue with now there's a negative on the left preventing me logging?

 

[Mark44]

- l/12.5-1=e^-t/cr

However doesn't this cause another issue with now there's a negative on the left preventing me logging?

i is a variable - it could be positive or negative.

In your original equation, which was $i = 25(1 - e^{-t/(cr)})$, if e^-t/(cr) > 1, then i will be negative.

 

[Ramjam]

i is a variable - it could be positive or negative.

In your original equation, which was $i = 25(1 - e^{-t/(cr)})$, if e^-t/(cr) > 1, then i will be negative.

The original equation was i = 12.5(1-e^-t/cr the reason it became i over 12.5 was because i divided 12.5 on both sides to remove it from the right.

 

[Ramjam]

Heres the full question that i should of probably started with (my mistake) The current, i, flowing in a capacitor at time, t, is given by i = 12.5(1-e^-t/cr where resistance, R, is 30KΩ and capacitance, C, is 20μF Determine the time for the current to reach 10 amperes?

So i know that by dividing both sides by 12.5 allows me to get i/12.5 = 1-e^-t/cr and then by subtracting -1 from both sides gives me i/12.5 -1 = -e^-t/cr, then by multiplying by -1 give me - i/12.5 = e^-t/cr from this point on can i log both sides as the minus on the left is there?

 

[Mark44]

Heres the full question that i should of probably started with (my mistake) The current, i, flowing in a capacitor at time, t, is given by i = 12.5(1-e^-t/cr where resistance, R, is 30KΩ and capacitance, C, is 20μF Determine the time for the current to reach 10 amperes?

Yes, you should have given us the complete problem statement. With the i, C and R, I figured this was a problem about an RC circuit.

So i know that by dividing both sides by 12.5 allows me to get i/12.5 = 1-e^-t/cr and then by subtracting -1 from both sides gives me i/12.5 -1 = -e^-t/cr, then by multiplying by -1 give me - i/12.5 = e^-t/cr

No. Before you multiplied, the left side was i/12.5 - 1. After you multiplied by -1, what happened to the -1?

from this point on can i log both sides as the minus on the left is there?

 

[Ramjam]

opps sorry that should be -i/12.5-1=e^-t/cr

 

[Mark44]

opps sorry that should be -i/12.5-1=e^-t/cr

No, that's not right either. You really need to be more careful.

 

[Ramjam]

in that case i don't under what you mean by

After you multiplied by -1, what happened to the -1?

 

[Mark44]

What is -1 times i/12.5 - 1?

If you're having such a difficult time with basic algebra, I am very concerned about the prospects of you working with exponential and log equations.

 

[Ramjam]

What is -1 times i/12.5 - 1?

If you're having such a difficult time with basic algebra, I am very concerned about the prospects of you working with exponential and log equations.

It was 2am in the uk and after a long working day which i also do along side my course probably wasn't the best idea.

Right here are my working so far now that I've been had some sleep,

we start with I=12.5(1-e^-t/cr)

I then divide both sides by 12.5 removing the 12.5 from the right hand side.

I/12.5 = 1-e^-t/cr

I then subtract -1 from both sides

I/12.5 -1 = -e^-t/cr

I then times both sides by -1 giving me

(1- I/12.5) = e^-t/cr

how ever to remove I am unsure of how to go on from here?

 

[ehild]

You need the time when the current I reaches the value I=10 A. Note that 12.5 is also in amperes. Substitute 10 for I and evaluate the left side of the equation.
Also calculate cr, or better CR from the given value of capacitance C and resistance R.