Isolating Exponent Variables for Deriving Equations

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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

$$A_1 = fracA_nr^n-1 $$ $$ r = sqrt[n-1]frac A_nA_1 $$
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

sequences-and-series derivatives exponentiation

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asked Jul 20 at 9:36

Dudamesh 192

235

add a comment | 

up vote
1
down vote

favorite

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

$$A_1 = fracA_nr^n-1 $$ $$ r = sqrt[n-1]frac A_nA_1 $$
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

sequences-and-series derivatives exponentiation

share|cite|improve this question

asked Jul 20 at 9:36

Dudamesh 192

235

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

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

$$A_1 = fracA_nr^n-1 $$ $$ r = sqrt[n-1]frac A_nA_1 $$
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

sequences-and-series derivatives exponentiation

share|cite|improve this question

asked Jul 20 at 9:36

Dudamesh 192

235

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

$$A_1 = fracA_nr^n-1 $$ $$ r = sqrt[n-1]frac A_nA_1 $$
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

sequences-and-series derivatives exponentiation

share|cite|improve this question

asked Jul 20 at 9:36

Dudamesh 192

235

share|cite|improve this question

share|cite|improve this question

asked Jul 20 at 9:36

Dudamesh 192

235

asked Jul 20 at 9:36

Dudamesh 192

235

asked Jul 20 at 9:36

Dudamesh 192

235

235

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
1
down vote



accepted

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 $fraclog(729)log(3)$ 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

$fracA_nA_1 = r^n-1$

Apply log() to both sides

$log(fracA_nA_1) = log(r^n-1)$

Apply that law of log about exponents

$log(fracA_nA_1) = (n-1)cdot log(r)$

Divide both sides by $log(r)$

$fraclog(fracA_nA_1)log(r) = n-1$

Add one to both sides

$$n = fraclog(fracA_nA_1)log(r)+1$$

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

share|cite|improve this answer

edited Jul 29 at 0:42

answered Jul 20 at 10:18

Dudamesh 192

235

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your workings are correct :) You can simplify the last expression a bit more to $$n=fraclg A_n - lg A_1lg r+1$$
  – Karn Watcharasupat
  Jul 21 at 3:22
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

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 $fraclog(729)log(3)$ 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

$fracA_nA_1 = r^n-1$

Apply log() to both sides

$log(fracA_nA_1) = log(r^n-1)$

Apply that law of log about exponents

$log(fracA_nA_1) = (n-1)cdot log(r)$

Divide both sides by $log(r)$

$fraclog(fracA_nA_1)log(r) = n-1$

Add one to both sides

$$n = fraclog(fracA_nA_1)log(r)+1$$

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

share|cite|improve this answer

edited Jul 29 at 0:42

answered Jul 20 at 10:18

Dudamesh 192

235

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your workings are correct :) You can simplify the last expression a bit more to $$n=fraclg A_n - lg A_1lg r+1$$
  – Karn Watcharasupat
  Jul 21 at 3:22
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

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 $fraclog(729)log(3)$ 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

$fracA_nA_1 = r^n-1$

Apply log() to both sides

$log(fracA_nA_1) = log(r^n-1)$

Apply that law of log about exponents

$log(fracA_nA_1) = (n-1)cdot log(r)$

Divide both sides by $log(r)$

$fraclog(fracA_nA_1)log(r) = n-1$

Add one to both sides

$$n = fraclog(fracA_nA_1)log(r)+1$$

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

share|cite|improve this answer

edited Jul 29 at 0:42

answered Jul 20 at 10:18

Dudamesh 192

235

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your workings are correct :) You can simplify the last expression a bit more to $$n=fraclg A_n - lg A_1lg r+1$$
  – Karn Watcharasupat
  Jul 21 at 3:22
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

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 $fraclog(729)log(3)$ 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

$fracA_nA_1 = r^n-1$

Apply log() to both sides

$log(fracA_nA_1) = log(r^n-1)$

Apply that law of log about exponents

$log(fracA_nA_1) = (n-1)cdot log(r)$

Divide both sides by $log(r)$

$fraclog(fracA_nA_1)log(r) = n-1$

Add one to both sides

$$n = fraclog(fracA_nA_1)log(r)+1$$

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

share|cite|improve this answer

edited Jul 29 at 0:42

answered Jul 20 at 10:18

Dudamesh 192

235

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 $fraclog(729)log(3)$ 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

$fracA_nA_1 = r^n-1$

Apply log() to both sides

$log(fracA_nA_1) = log(r^n-1)$

Apply that law of log about exponents

$log(fracA_nA_1) = (n-1)cdot log(r)$

Divide both sides by $log(r)$

$fraclog(fracA_nA_1)log(r) = n-1$

Add one to both sides

$$n = fraclog(fracA_nA_1)log(r)+1$$

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

share|cite|improve this answer

edited Jul 29 at 0:42

answered Jul 20 at 10:18

Dudamesh 192

235

share|cite|improve this answer

share|cite|improve this answer

edited Jul 29 at 0:42

edited Jul 29 at 0:42

edited Jul 29 at 0:42

answered Jul 20 at 10:18

Dudamesh 192

235

answered Jul 20 at 10:18

Dudamesh 192

235

answered Jul 20 at 10:18

Dudamesh 192

235

235

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your workings are correct :) You can simplify the last expression a bit more to $$n=fraclg A_n - lg A_1lg r+1$$
  – Karn Watcharasupat
  Jul 21 at 3:22
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your workings are correct :) You can simplify the last expression a bit more to $$n=fraclg A_n - lg A_1lg r+1$$
  – Karn Watcharasupat
  Jul 21 at 3:22
  

  
  
  
  

Your workings are correct :) You can simplify the last expression a bit more to $$n=fraclg A_n - lg A_1lg r+1$$
– Karn Watcharasupat
Jul 21 at 3:22

Your workings are correct :) You can simplify the last expression a bit more to $$n=fraclg A_n - lg A_1lg r+1$$
– Karn Watcharasupat
Jul 21 at 3:22

add a comment | 

 

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