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Our Math class right now is all about Geometric Sequences

and we all know the equation for them which is

$$A_n = A_1 \cdot r^{n-1}$$

We were only given this particular equation so to make things faster for myself , I derived some equations for specific variables

$$A_1 = \frac{A_n}{r^{n-1}} $$ $$ r = \sqrt[n-1]{\frac {A_n}{A_1}} $$ However, after a couple of days of thinking I cannot seem to get a proper derived equation for the variable n in that particular equation. In the end I resorted to this method where I manually think up of an exponent and base... for Example:

$A_n = 2916$

$A_1 = 4$

$r = 3$

$2916 = 4 \cdot3^{n-1} $

isolating $r^{n-1}$ so we divide 4 to both sides cancelling 4 and 2916 will become 729

$729 = 3^{n-1} $

using 3 as a base, 729 can be shortened to $3^6$

$3^6 = 3^{n-1} $

using simple logic we know that n = 7


But enough context, on the formula above, I want to know how exactly do I isolate the n variable like how I can divide both sides to 'transpose' a variable being multiplied to another term

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After researching a bit I found out the existence of logarithms which are apparently the inverse of exponents, the same as how divide is the inverse of multiply.

After studying logarithm laws for a good hour I found out another way to finish the same question:

$2916 = 4 \cdot3^{n-1}$

divide 2916 by 4 to isolate $3^{n-1}$

$729 = 3^{n-1}$

Apply log() to both sides

$log(729) = log(3^{n-1})$

one law of log states that $log(x^y)$ becomes $y \cdot log(x)$ so..

$log(729) = (n-1)\cdot log(3)$

then divide log(3) on both sides, that means $\frac{log(729)}{log(3)}$ which is equal to 6

$ 6 = n -1 $ or simply $n = 7$


Correct me if im wrong but this is my attempt at deriving the original equation to a one-line equation

$A_n = A_1 \cdot r^{n-1}$

divide $A_1$ on both sides

$\frac{A_n}{A_1} = r^{n-1}$

Apply log() to both sides

$log(\frac{A_n}{A_1}) = log(r^{n-1})$

Apply that law of log about exponents

$log(\frac{A_n}{A_1}) = (n-1)\cdot log(r)$

Divide both sides by $log(r)$

$\frac{log(\frac{A_n}{A_1})}{log(r)} = n-1$

Add one to both sides

$$n = \frac{log(\frac{A_n}{A_1})}{log(r)}+1$$

Hopefully I answered my own question, if I did, well... might as well keep this here for future reference

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  • $\begingroup$ Your workings are correct :) You can simplify the last expression a bit more to $$n=\frac{\lg A_n - \lg A_1}{\lg r}+1$$ $\endgroup$ – Karn Watcharasupat Jul 21 '18 at 3:22

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