A little bit of help rearranging and equation

[SMOF]

Hello,

Hopefully someone can help me rearrange this equation for x².

The equation is

C_B = C_o e^(-x2/4Dt)

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

Thanks.

Seán

 

[gb7nash]

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

 

[SMOF]

Hello.

Well, that would be

C_B/C_o = e^(-x2/Dt)

 

[gb7nash]

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

 

[SMOF]

See, this is where I am not sure.

Is it

ln(C_B/C_o) = ln(-x²/Dt)

?

 

[gb7nash]

See, this is where I am not sure.

Is it

ln(C_B/C_o) = ln(-x²/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?

 

[SMOF]

So,

ln(C_B/C_o) = ln(e^(-x2/Dt))

?

 

[gb7nash]

That's fine. One well-known rule of ln is:

ln(a^b) = b ln(a)

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

 

[SMOF]

Well, that would be

ln(C_B/C_o) = (-x²/Dt)ln(e)

and ln(e) is 1?

 

[gb7nash]

Well, that would be

ln(C_B/C_o) = (-x²/Dt)ln(e)

and ln(e) is 1?

Correct. You can probably take it from here.

 

[SMOF]

Yea, that's fantastic!

Many thanks!

Seán