I Logarithmic scale - interpolation

AI Thread Summary
To interpolate on a logarithmic scale, the process involves transforming the coordinates of the points into logarithmic values. The formula for linear interpolation can be adapted to logarithmic scales by replacing y values with log(y) and x values with log(x). This allows for the use of the same interpolation method, but the results will be in logarithmic form. After calculating log(y), the original y value can be recovered using the exponential function, y = exp(log(y)). This method effectively allows for accurate interpolation between points on a log-log plot.
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How can I interpolate between two values on a logarithmic scale plot?
Hi,

knowing the coordinates of two points: ##(x_1,y_1)## and ##(x_2,y_2)## on a linear scale plot, I can use linear interpolation to get ##y## for a point of known ##x## using the formula below: $$y=y_1+(x−x_1) \frac{(y_2−y_1)}{(x_2−x_1)}$$
But how does it look like in the case of logarithmic scale (log-log plot)? How can I get ##y## for known ##x## when I have the coordinates of two other points? So far I haven't found any working formula for that.

Thanks in advance for your help.
 
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FEAnalyst said:
when I have the coordinates of two other points?
You have to specify what you mean by that ! Example: Plot ##y = x^3## on log-log paper. Suppose you want to interpolate between ##(3,27)## and ##(4,64)## to find ##3.75^3## (is 52.73).

1619110562193.png


Do you have the coordinates of those points as found on the axis, or in mm on the paper ?

In the first case your $$y-y1=(x−x_1) \frac{(y_2−y_1)}{(x_2−x_1)}$$is still 'valid' in the logarithm world$$
\log{y\over y_1} = \log{x\over x_1}*{\log(y_2/y_1)\over \log(x_2/x_1)}$$as you can easily check with a calculator (or excel ?:) ):
1619109772500.png


In the second case you do something similar, but you already have the logarithms.
1619110493181.png


##\ ##
 
Just replace all y values by log(y) and all x-values by log(x). That's all. That's exactly what a log-log plot does. Your result is then log(y), but of course you can recover y using y = exp(log(y)).
 
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