 # What is the difference between logarithmic decay vs exponential decay? I aangenaamyj 2022-07-09 Answered
What is the difference between logarithmic decay vs exponential decay?
I am a little unclear on whether they are distinctly different or whether this is a 'square is a rectangle, but rectangle is not necessarily a square' type of relationship.
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The "Square is a rectangle" relationship is an example where the square is a special case of a rectangle.
"Exponential decay" gets its name because the functions used to model it are of the form $f\left(x\right)=A{e}^{kx}+C$ where $A>0$ and $k<0$. (Other k's above 0 yield an increasing function, not a decaying one.)
Similarly for "logarithmic decay," it gets its name since its modeled with functions of the form $g\left(x\right)=A\mathrm{ln}\left(x\right)+C$ where $A<0$
These two families of functions do not overlap, so neither is a special case of the other. The giveaway is that the functions with $\mathrm{ln}\left(x\right)$ aren't even defined on half the real line, whereas the exponential ones are defined everywhere.

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The natural logarithm and exponential are inverses of one another, so the associated slopes will also be inverses. If you put exponentially decaying data on a log plot, i.e. log of the exponential decaying data with the same input, you get a linear plot. If you put the logarithmic decaying plot on an exponential plot (exponential of the data), you get a linear plot, so the way they are decaying is exactly opposite.

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