- #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

- 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

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

- #5

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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

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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?See, this is where im not sure.

Is it

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

?

- #7

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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

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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

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Correct. You can probably take it from here.Well, that would be

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

and ln(e) is 1?

- #11

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Yea, thats fantastic!

Many thanks!

Seán

Many thanks!

Seán

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