Struggling to rearrange a formula to find c

[leejohnson222]

Homework Statement

Vc = Vs ( 1 -e -t/rc )
i can plug the numbers in to a point but im unsure about rearranging to create the new formula to find c

Relevant Equations

Vc = 1v Vs = 22v t= 3

I am trying to get to grips with rearranging this formula to find c
Vc = Vs ( 1 -e -t/rc )
i can plug the numbers in to a point but im unsure about rearranging to create the new formula

 

[haruspex]

Homework Statement:: Vc = Vs ( 1 -e -t/rc )
i can plug the numbers in to a point but im unsure about rearranging to create the new formula
Relevant Equations:: Vc = 1v Vs = 22v t= 3

I am trying to get to grips with rearranging this formula to find T
Vc = Vs ( 1 -e -t/rc )
i can plug the numbers in to a point but im unsure about rearranging to create the new formula

I assume you mean $V_c = V_s ( 1 -e ^{-\frac t{rc}} )$.
If you were to calculate V_c from it you would start by dividing t by r and c, and finish by multiplying the (evaluated) term in parentheses by V_s.
Just reverse that sequence. Divide V_c by V_s etc.

 

[leejohnson222]

yes i did mean that, so would it be 1- Vc/Vs = e - t/rc ? i need to understand each step so i can apply this to other problems.

 

[haruspex]

yes i did mean that, so would it be 1- Vc/Vs = e - t/rc ?

Yes, but keep going. How do you 'undo' the exponentiation?

 

[leejohnson222]

is this where log comes in ? and you need to do apply log to both sides of the equation?

 

[haruspex]

is this where log comes in ? and you need to do apply log to both sides of the equation?

Yes, you must always do the same thing to each side of an equation if you want to be sure it is still true.

 

[leejohnson222]

ok this is where i am a little stuck in writing this out, from 1 - vc/vs = e - t/rc
i understand i need to undo e but its expressing it and breaking down the formula again
log e 1 - vc/vs log e - t/rc that cant be correct

 

[haruspex]

ok this is where i am a little stuck in writing this out, from 1 - vc/vs = e - t/rc
i understand i need to undo e but its expressing it and breaking down the formula again
log e 1 - vc/vs log e - t/rc that cant be correct

You put "log e" in front on the left but only "log" in front on the right.
You need to be more exact about how you write equations.
You had $1 - v_c/v_s = e ^{- t/rc}$.

Taking logs both sides, you need parentheses :
$\ln(1 - v_c/v_s) = \ln(e ^{- t/rc})$.
Note that ln () means log to the base e.
If LaTeX is beyond you, at least use subscripts and superscripts. These are the X₁ and X¹ buttons.

What is $\ln(e^x)$?

 

[leejohnson222]

ok sure i see what you mean i think, are you asking me to rearrange this formula, i dont know how to write these correctly on this forum?

 

[leejohnson222]

is this loge (e^x)=x ? i think im getting mixed up, i am not sure where to take both sides of this equation now or how to translate the numbers in

 

[kuruman]

Yes, $\log_e(e^x)=x$.

 

[leejohnson222]

i am still not confident i can break the formula down from its current point

 

[kuruman]

i am still not confident i can break the formula down from its current point

Let's summarize what we have so far.

1. $\ln(1 - v_c/v_s) = \ln(e ^{- t/rc})$
2. $\ln(e^x)=x$

Therefore $$ \ln(e ^{- t/rc})=?$$

 

[leejohnson222]

sorry hit a wall with this now, i dont see how its breaking down to find c
i think im confused with x in the formula now

 

[willem2]

ok this is where i am a little stuck in writing this out, from 1 - vc/vs = e - t/rc
i understand i need to undo e but its expressing it and breaking down the formula again
log e 1 - vc/vs log e - t/rc that cant be correct

If it's too hard to use Latex for equations, you should use a least
1 - Vc/Vs = e ^-t/rc (select -t/rc, use the supersciript button)
I don't see what the problem is with taking the log of both sides. The right side of the equaton will be the log of an exponential, so you should be able to simplify that.

 

[Herman Trivilino]

i think im confused with x in the formula now

Replace $x$ with $-\frac{t}{RC}$.

 

[leejohnson222]

If it's too hard to use Latex for equations, you should use a least
1 - Vc/Vs = e ^-t/rc (select -t/rc, use the supersciript button)
I don't see what the problem is with taking the log of both sides. The right side of the equaton will be the log of an exponential, so you should be able to simplify that.

ok so In (1 - 1/22) = In ( e^-3/1xc )
log (1 - 0.0454) =<sup>-3/c

-3/c</sup> = In (0.6667)

 

[leejohnson222]

Let's summarize what we have so far.

1. $\ln(1 - v_c/v_s) = \ln(e ^{- t/rc})$
2. $\ln(e^x)=x$

Therefore $$ \ln(e ^{- t/rc})=?$$

In ( 1 - Vc/vs ) ??

 

[epenguin]

I don't see what the problem is with taking the log of both sides. The right side of the equaton will be the log of an exponential, so you should be able to simplify that.

The problem seems to be he does not know or remember what a logarithm to a given base is, nor possibly what raising to a power is, what $e$ is, what $ln$ is, so you need to decide whether it is appropriate to for you to explain these things, or instead refer him to his old textbook.

 

[leejohnson222]

appologies i am new to this and was getting a hold of it to a degree, i understand log is another way to express the exp, e is the natural log In. I believe that if its there is a number after log thats the base, if not it can be assumed to be 10

 

[epenguin]

Well if there is one thing I learned here it is respect for the writers of textbooks, their job is a difficult one. I don't think we should duplicate it much. rather our job is to clear up the misunderstandings and oversights, points missed that still emerge afterwards. Though go ahead anyone who wants to try.

But I think you would be best off by looking back at your old math textbook, or equivalent on the web, you'd probably find it quite soon comes back to you.

 

[Herman Trivilino]

i understand log is another way to express the exp

No. They are inverse functions. Hence $\ln{e^x}=x$.

 

[haruspex]

No. They are inverse functions. Hence $\ln{e^x}=x$.

… and $e^{\ln(x)}=x$.