Graphing Log-Linear Analysis of 2 Exponential Functions

[AlexC]

I need to graph a relationship that is the combination of 2 exponential functions on log-linear graph paper.

I have: V = Cexp(-t/B) + Aexp(-t/S)

If it were simply V = Cexp(-t/B) then this wouldn't be a problem (by graphing log(V) = (-1/B)log(e)t + log(C) ) but I need to determine both the B and S parameters (possibly from the same graph). How can i do this?

 

[gammamcc]

If t=ln(s), then V is linear in s...

Plot V vs. s.

 

[tacman]

Graphing log-linear analysis of two exponential functions can be a bit challenging, but it is definitely possible. First, let's break down the equation V = Cexp(-t/B) + Aexp(-t/S) to understand what each of the variables represents.

V represents the dependent variable, which is the output or value we are trying to find. In this case, it could represent something like population size or growth rate.

C and A represent the initial values or starting points for each of the exponential functions. These values determine the y-intercepts on the graph.

B and S represent the growth or decay rates for each of the functions. These values determine the slope of the graph.

To graph this relationship, we will use log-linear graph paper. This type of graph paper has a logarithmic scale on the x-axis and a linear scale on the y-axis. This allows us to plot exponential functions on a linear graph.

To start, we will take the natural log of both sides of the equation:

ln(V) = ln(Cexp(-t/B) + Aexp(-t/S))

Using the properties of logarithms, we can rewrite this as:

ln(V) = ln(C) + ln(exp(-t/B)) + ln(A) + ln(exp(-t/S))

Now, we can simplify further by using the rules of exponents:

ln(V) = ln(C) + (-t/B)ln(e) + ln(A) + (-t/S)ln(e)

Since ln(e) = 1, we can simplify even further to get:

ln(V) = ln(C) - t/B + ln(A) - t/S

Now, we have an equation in the form of y = mx + b, where y = ln(V), m = -1/B and b = ln(C) + ln(A) - t/S.

On the log-linear graph paper, we will plot ln(V) on the y-axis and t on the x-axis. The slope of the line will be -1/B and the y-intercept will be ln(C) + ln(A) - t/S.

Next, we will repeat this process for the second exponential function:

ln(V) = ln(Cexp(-t/B) + Aexp(-t/S))

ln(V) = ln(C) + ln(exp(-t/B)) + ln(A) + ln(exp(-t/S))

ln(V) = ln