Can anyone solve how long it would take for investment A to equal investment B.?

Question

An investor buying investment A pays a front end charge, Investment B has no front end charge but has higher ongoing costs. I need to solve for n. n needs to be a natural number.

n = number or periods
LD = front end charge
r = rate of return
b = the higher ongoing cost in investment B over investment A.

The return for investment a should be (1- LD)/ (1+r)^n
The return for B should be 1/(1+(r-b))^n

I need the value of n when A = B. I was given a formula, but n ended up negative. That doesn't help.

A correction to my formula was posted.
FV_A = (1-LD)*(1+r)^n = (1+(r-b))^n

A correction to my formula was posted.
FV_A = (1-LD)*(1+r)^n = (1+(r-b))^n

Answer 1

(1- LD)/ (1+r)^n = 1/(1+(r-b))^n

cross multiply...

(1 - LD)(1 + (r-b))^n = 1 (1 + r)^n

Take the natural log of both sides:

ln [(1 - LD)(1 + r - b)^n] = ln (1 + r)^n
ln (1 - LD) + ln (1 + r - b)^n = ln (1 + r)^n
ln (1 - LD) + n*ln (1 + r - b) = n * ln (1 + r)
n*ln(1 + r - b) - n*ln(1 + r) = - ln (1 - LD)
n[ln (1 + r - b) - ln (1 + r)] = -ln (1 - LD)
n = -ln(1 - LD) / [ln (1 + r - b) - ln (1 + r)]

Yes, it looks like n is a negative number (because of the -ln... on the top of the fraction). However, once you plug numbers in for all those variables, that may not be the case. I can't tell because I don't know what the variables would be.

Answer 2

(1- LD)/ (1+r)^n = 1/(1+(r-b))^n
ln(1-LD) - nln(1+r) = ln1 - nln (1+r-b)
nln(1+r-b) -nln(1+r) = 0 -ln(1-LD)
n(ln(1+r-b) - ln(1+r) = -ln(1-LD)
n = -ln(1-LD)/[(ln(1+r-b) - ln(1+r)]

This can't be the right answer since 1-LD will be negative and so ln(1-LD) will be undefined.

Are you sure your two equations for return on investment is correct. It appears that the return for A will always be negative because 1-LD will always be negative and (1+r)^n will always be positive so their quotient will always be negative. A would be a very poor investment

In B, I don't understand why you are subtracting the ongoing cost from the rate of return. That can't be right.

Answer 3

From your Question we have
If A = (1- LD)/ (1+r)^n
and B = 1/(1+(r-b))^n
and If A = B then

(1- LD)/ (1+r)^n = 1/(1+(r-b))^n
or taking log on both sides we get
log(1-LD) - nlog(1+r) = log(1) - nlog(1+(r-b))
or log(1-LD)=n[log(1+r)-log(1+(r-b))]
or n = log(1-LD)/[log(1+r)-log(1+r-b)]
or n = log(1-LD)/log[(1+r)/(1+r-b)]

Answer 4

You have the formulae wrong.

FV_A = (1-LD)*(1+r)^n <---- multiply, don't divide
FV_B = (1+(r-b))^n <----- not the reciprocal

Answer 5

(1- LD)/ (1+r)^n = 1/(1+(r-b))^n
ln(1-LD) - nln(1+r) = ln1 - nln (1+r-b)
nln(1+r-b) -nln(1+r) = 0 -ln(1-LD)
n(ln(1+r-b) - ln(1+r) = -ln(1-LD)
n = -ln(1-LD)/[(ln(1+r-b) - ln(1+r)]

This can't be the right answer since 1-LD will be negative and so ln(1-LD) will be undefined.