Rearranging x=x0e^-lambda t in the form y=mx+c

[Mike Shandon]

Hi, How could I rearrange x=x0e^-lambda t into the form y=mx+c, where y is equal to ln x and x is equal to t?

Thank you in advance

I tried to solve the problem myself, by taking the natural log of both sides, this left me with:

ln x = -lambda t * lnx0

However, I am not sure if this answer is correct or not

If this is correct, would -lambda represent the gradient? And, would ln x0 represent the y intercept

 

[mfb]

What is h0?
The logarithm of a product is the sum of logarithms, so you should get $-\lambda t + \ln(x_0)$.

If this is correct, would -lambda represent the gradient? And, would ln h0 represent the y intercept

Sure.

 

[Mike Shandon]

Thank you

 

[Mike Shandon]

What is h0?
The logarithm of a product is the sum of logarithms, so you should get $-\lambda t + \ln(x_0)$.
Sure.

My mistake h0 was supposed to represent x0