# interpolating on logarithmic graphs

• September 17th 2010, 09:08 PM
TheBerkeleyBoss
interpolating on logarithmic graphs
Hi everyone,

I am taking Properties of Materials, however, there is a question that involves some applied algebra.

Here is my problem: I am shown a graph of a line with a positive slope, however, the abscissa is in logarithmic scale. Even though the graph appears to be linear, it says the function is NOT linear. It wants me to interpolate data, but using y = mx + b does not give me consistent answers (even though I have what appears to be a y-intercept and linear-looking line). How can I interpolate data on logarithmic scales? I believe this would be considered a semi-logarithmic plot.

Thank you!
• September 18th 2010, 03:31 AM
HallsofIvy
Well, first, y= mx+ b does not give "consistent answers" because the function is not linear! Any linear function must be of the form for some x and y. Also "interpolation" (I believe you mean "linear interpolation"- there are many different kinds of interpolation) is an approximation method, it does not give "the" correct value. You interpolate between two given values by assuming a straight line graph between those two points. Of course, if your graph is NOT a straight line, that will not give you the correct answer- but it should give an approximately correct answer. If $(x_0, y_0)$ and $(x_1, y_1)$ are points on either side of your desired x value, then the straight line through them is given, using the "slope-intercept form" by $y= \left(\frac{y_1- y_0}{x_1- x_0}(x- x_0)+ y_0$. You can also think of that (by a little shifting of the fractions) $y= \frac{x- x_0}{x_1- x_0}(y_1- y_0)+ y_0$ where you are determining what fraction of the way x is between $x_0$ and $x_0$ and taking that fraction of the distance between $y_0$ and $y_1$ and adding it to $y_0$.

If I understand this correctly, your "y" axis is actually giving ln(y) so that your graph is really ln(y)= f(x) and so $y= e^{f(x)}$. If the line, in the "semi-logarithmic plot", is actually straight then you have ln(y)= ax+ b so $y= e^{ax+ b}$.