Hello,

In the following formula, I want to determine "n" which is an exponent

X - [(p)(a)(i^n)] = 0
X = (p)(a)[(1+i)^n]

I tried to use the seriessum function, but "n" is a required input. Which
function do I use to solve for n? Thank you.

When solving for an exponent, take the log of both sides.

X = (p)(a) (i^n)
X/pa=i^n
log (X/pa)=n log (i)
n=log(x/pa)/log(i)

You get the idea.

"sirsoto" wrote:

Hello,

In the following formula, I want to determine "n" which is an exponent

X - [(p)(a)(i^n)] = 0
X = (p)(a)[(1+i)^n]

I tried to use the seriessum function, but "n" is a required input. Which
function do I use to solve for n? Thank you.

Thanks Barb,

I tried Logarithm but it did not work because (1+i) increases annually by
5%. My formala for example looks like this:

X - [(p)(a)(i^n)] = 0
Year 1: X0 - [(p)(a)[(1+i)^0] = X1
Year 2: X1- [(p)(a)[(1+i)^1] = X2
Year 3: X2- [(p)(a)[(1+i)^2] = X3

and so on until X0 (from Year 1) equals zero. This is sort of like
break-even.

Regards, SirSoto


"Barb R." wrote:

When solving for an exponent, take the log of both sides.

X = (p)(a) (i^n)
X/pa=i^n
log (X/pa)=n log (i)
n=log(x/pa)/log(i)

You get the idea.

"sirsoto" wrote:

Hello,

In the following formula, I want to determine "n" which is an exponent

X - [(p)(a)(i^n)] = 0
X = (p)(a)[(1+i)^n]

I tried to use the seriessum function, but "n" is a required input. Which
function do I use to solve for n? Thank you.