# I can't find (lny-lny0)/t meaning [closed]

I cannot understand what (lny-lny0)/t means. Of course we use this to show growth rate. But in the mathematical sense, I cannot know what it is. Is it diferrentiaion?

• Hi and welcome tk Economics Stack Exchange. Can you please give us some more context. Where did you find the expression and what was it used for there? – BB King Sep 22 '18 at 15:37
• to find out growth rate, first, y=y0(1+g)^t and lny=lny0+tln(1+g) and (lny-lny0)/t = ln(1+g) using maclaurin expansion ln(1+g) = g so (lny-lny0)/t = g – user19506 Sep 22 '18 at 15:39
• Growth rate of what? What is t, what is g, what is y? How did the author you saw this from explain it? Please add the info in your comment and all other info to the main text of the question. – BB King Sep 22 '18 at 15:40
• Oh i edit my explanation. – user19506 Sep 22 '18 at 15:44
• It would help if you edited the question using the math markup that is available on this site, and to maybe write out some of these formulas you have in the comments up there. – Kitsune Cavalry Sep 22 '18 at 17:31

Generally ln is used to represent the natural logarithm. I know of no notation system that uses it to represent differentiation.

Natural logarithm:

$$\frac{\ln{y} - \ln{y_0}}{t}$$

Differentiation would look more like

$$\frac{dy}{dt}$$

or

$$f'(t)$$

It is common to take logarithms of growth functions because they involve exponents and logarithms can simplify equations with exponents. In particular,

$$\ln{(1+g)^t} = t\ln{(1+g)}$$

is a basic property of logarithms.

A Maclaurin expansion is an alternative way to write the equation. See Taylor series and Maclaurin's series for more information. A Maclaurin expansion does in fact take advantage of differentiation.

This logarithm is referred to as natural because its base arises from normal mathematical calculations rather than being arbitrary, much like $$\pi$$.

• Is there any intuition of using (lny-lny0)/t? – user19506 Sep 23 '18 at 2:08