converting an exponential to a straight line, why and how

Sometimes when you are dealing with an exponential graph it is just a whole lot easier to convert the exponential graph to a straight line graph (you have done this a couple of times already in 107).

For example, remember this graph - it shows how growth rate increases with increasing temperature, and the part of the graph between points a and b represents exponential increase in growth rate with increasing temperature.

Now remember this equation:

Where Rt2 is the rate of growth at temperature 2

Rt1 is the rate at growth at temperature 1

Q10 is the factor by which the rate increases for every 10 degrees increase in temperature

t2 is temperature 2

t1 is temperature 1

(and remember the dot is a short hand way of writing a times sign -'x')

If you want to calculate Q10, how are you going to do it? You could read a couple of points off the exponential graph and enter them into the equation above, and then re-arrange it to solve it for Q10. This is ok but not the best way - this is because in real life science not all of your points will have perfectly fitted the exponential curve you have drawn. This is because all of your points are not entirely accurate (there was error in your experiment!) and so using just two points to calculate Q10 is likely to give you a less accurate value for Q10 than if you were to use all of your data to help you calculate Q10.

In practise it is much better to convert the exponential equation to an equation that describes a straight line graph. This way you can re-plot your data and easily find Q10 using all of your data (which will be the gradient of the straight line). Let me show you how you can do this.....

Your starting point is your exponential equation:

                                                  R_t2 = R_t1 x Q10 ^[(t2-t1)/10]

First you have to log everything on both sides of the equation:

logR_t2 = log (R_t1 x Q10 ^[(t2-t1)/10] )

Now you can simplify the right side using your laws of how to manipulate logarithms (see my previous post on laws of logarithms):

Using rule 1:  Log₁₀ (m x n)    =    Log₁₀m + Log₁₀n :

logR_t2 = logR_t1 + logQ10 ^[(t2-t1)/10]

Using rule 3: Log₁₀ (mⁿ)    =    n x Log₁₀m  :

                                   logR_t2 = logR_t1 + [(t2 - t1)/10] x logQ10 

I know it might not seem obvious, but this is now an equation for a straight line!!! - If you plot logR_t2 on the y axis and [(t2 - t1)/10] on the x axis you will get a straight line with gradient = logQ10 and y intercept = logR_t1 ...

                                   logR_t2 = logR_t1 + [(t2 - t1)/10] x logQ10

                                       y       =     c      +          x                m

Or if you rearrange the equation a little so it looks a little more familiar:

                                    logR_t2 =  logQ10  x   [(t2 - t1)/10]   +  logR_t1 

                                        y       =          m                x               +      c

So now to calculate Q10 you can use all of your data to draw a graph with logR on the y axis and t/10 on the x axis, and you can draw a line of best fit through your points -  

logQ10 will be he gradient of this line, and so you can do the antilog of this number to find the value for Q10!

phew