Undergrad Logarithmic scale - interpolation

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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.
FEAnalyst
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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)).
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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