Posted by **Jackie** on Friday, January 25, 2013 at 12:39pm.

a) 1150 dollars invested at 9% annual interest rate (compounded yearly) or

b) 1475 invested at 6% annual interest (compounded yearly) after

When would the two investments have equal value?

- math -
**Reiny**, Friday, January 25, 2013 at 1:34pm
let the time be t

1150(1.09)^t = 1475(1.06)^t

46(1.09)^t = 59(1.06)^t

take log of both sides , and using simple log rules

log46 + tlog1.09 = log59 + tlog1.06

tlog1.09 - tlog1.06 = log 59 - lot 46

t(log 1.09 - log 1.06) = log 59 - log 46

t = (log59 -log46)/(log1.09 - log1.06) = 8.918

it would take appr 9 years

check:

1150(1.09^9 = 2497.68

1475(1.06)9 = 2491.98

using t = 8.918

1150(1.09)^8.918 = 2480.09

1475(1.06)^8.918 = 2480.13 , not bad

## Answer This Question

## Related Questions

- Math - melissa invested a some of money at 3% annual simple inter she invested ...
- algebra - Larry Mitchel invested part of his $32,000 advance at 7% annual simple...
- math - This problem has to do with exponential models. The question says, you ...
- algebra - A total of $12,000 is invested in two funds paying 9% and 11% simple ...
- college math - Suppose $400 is invested for 4 years at a nominal yearly interest...
- college math - Suppose $400 is invested for 4 years at a nominal yearly interest...
- math/algebra - Compounded semiannually. P dollars is invested at annual interest...
- algebra - Compounded semiannually. P dollars is invested at annual interest rate...
- algebra - mary had 25000 dollars to invest. she invested part of that amount at ...
- College Algebra - Compounded semiannually. P dollars is invested at annual ...

More Related Questions