- #1

- 74

- 0

Hopefully someone can help me rearrange this equation for x

^{2}.

The equation is

C

_{B}= C

_{o}e

^{(-x2/4Dt)}

It is changing the e into a log that is throwing me.

Thanks.

Seán

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- Thread starter SMOF
- Start date

- #1

- 74

- 0

Hopefully someone can help me rearrange this equation for x

The equation is

C

It is changing the e into a log that is throwing me.

Thanks.

Seán

- #2

gb7nash

Homework Helper

- 805

- 1

In order to apply the log, you need to isolate the e^{(-x2/4Dt)} term. How do you do that?

- #3

- 74

- 0

Hello.

Well, that would be

C_{B}/C_{o} = e^{(-x2/Dt)}

Well, that would be

C

- #4

gb7nash

Homework Helper

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- 1

Ok, now log both sides. What do you get?

- #5

- 74

- 0

See, this is where im not sure.

Is it

ln(C_{B}/C_{o}) = ln(-x^{2}/Dt)

?

Is it

ln(C

?

- #6

gb7nash

Homework Helper

- 805

- 1

See, this is where im not sure.

Is it

ln(C_{B}/C_{o}) = ln(-x^{2}/Dt)

?

The right-hand side is wrong. The left-hand side is fine. Take the ln of the right side, without trying to cancel anything out for now. What do we have?

- #7

- 74

- 0

So,

ln(C_{B}/C_{o}) = ln(e^{(-x2/Dt)})

?

ln(C

?

- #8

gb7nash

Homework Helper

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ln(a

How can we apply this to the right-hand side?

- #9

- 74

- 0

Well, that would be

ln(C_{B}/C_{o}) = (-x^{2}/Dt)ln(e)

and ln(e) is 1?

ln(C

and ln(e) is 1?

- #10

gb7nash

Homework Helper

- 805

- 1

Well, that would be

ln(C_{B}/C_{o}) = (-x^{2}/Dt)ln(e)

and ln(e) is 1?

Correct. You can probably take it from here.

- #11

- 74

- 0

Yea, thats fantastic!

Many thanks!

Seán

Many thanks!

Seán

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