SOLUTION: make n the subject if :s=a(1-r)^n/1-r

I had this wrong before, but Ikleyn pointed it out. So rather that leave
a wrong answer posted, I thought it would be better to correct it.

I'll bet you didn't mean this;


I'll bet you meant this



Nobody teaching algebra would intentionally put 1 as a denominator under a term
except in the most basic math problems.  Notice that the denominator on the
right is actually the same as (1-r) raised to the 1 power



So we can simply subtract the exponents of (1-r)



Take the natural logarithm of both sides:



Use rules for natural logs of a product on the right side;



Use the rule for the natural log of a power on the last term.





Use the distributive law on the right side. 



Isolate the only term containing n



Divide both sides of the equation by ln(1-r)



Isolate the term that contains n



Swap sides:



That is a correct and legitimate answer, but there are other
correct forms it could be changed to.

Edwin

The original formula is 

    s = .     (1)


In this formula, r must be different from 1 due to the Mathematical grammar rules; so, we accept this.


If so, we can reduce the factor (1-r) in the numerator and the denominator. We get then

    s = .    (2)


Next, we assume that a =/= 0.  

    (If a = 0, then s = 0, and 'n' can be any value; as such, it can not be determined from the formula).


Under this assumption, we can divide both sides of formula (2) by 'a'.  We get then

     = .     (3)


Now, the next natural assumption is that r < 1 and  > 0 

    (i.e., the ratio s/a is positive, or, in other words, real numbers 'a' and 's' are of the same sign).


Then we can safely take the logarithm of both sides in formula (3)

     = (n-1)*ln(1-r)


It gives 

    n-1 = ,

    n   =  + 1.



ANSWER.  Under natural assumptions r < 1, a =/= 0,  > 0,  the formula for 'n' is  

         n =  + 1.