Taking natural log of exponential

Question

I have the equation:

$$R=R_0e^{bx}$$

I'm given a table with values for $x$ and $R$, and asked to take natural logs of both sides and use a graph to find $b$.

So I think you rearrange the equation into the form $y=mx+c$, input the gradient and intercept after graphing the values of $x$ and $R$.

Here's what I've done so far:

$$\ln(R)=\ln(R_0e^{bx})\\ \ln(R)=\ln(R_0)+\ln(e^{bx})\\ \ln(R)=bx\,\ln(e)+\ln(R_0)\\ \ln(R)=bx+\ln(R_0)$$

So in this last line, is $\ln(R)$ the $y$, $bx$ the $mx$ term, and $\ln(R_0)$ the $c$ term?

Answer 1

If you have to rewrite your equation in terms of $\ln(x)$ and $\ln(R)$, as you mentioned in the title, this is the method:

$\ln(R)=\ln(R_0e^{bx})\\\ln(R)=\ln(R_0)+\ln(e^{bx})\\\ln(R)=\ln(R_0)+b\cdot e^{\ln(x)}$

$\to Y=b \cdot e^X+c$

where $Y=\ln(R)$, $X=\ln(x)$ and $c=\ln(R_0)$

Otherwise, your solution is ok. You are looking for $b$ so don't worry about other things as long as your solution is right and it is. You just need to find the slope of the line, that is equal to $b$.

Answer 2

Yes it is correct, now you need to plot your data $\ln(R)=bx+\ln(R_0)$ as a line of equation $y=mx+c$ assuming

• $y=\ln(R)$
• $x=x$
• $c=\ln(R_0)$

and you can find the angular coefficient $m=b$.

For that you can also use numerical methods as least square.