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

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1. Mar 13, 2016

### 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?

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

Last edited: Mar 13, 2016
2. Mar 13, 2016

### Staff: Mentor

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

3. Mar 13, 2016

### Mike Shandon

Thank you

4. Mar 13, 2016

### Mike Shandon

My mistake h0 was supposed to represent x0