Hello from MrBillDoesMath!
Answer: about 12 years ( approx 11.9 years)
Discussion:
Salary at ...
end of year 1: 25000 + (.06) * 25000
end of year 2: (salary at end of year 1) +
.06(salary at end of year 1) =
(25000 + (.06)*25000) +
.06 (25000 + (.06)*25000)
The bold faced terms are the same. Factor out the boldfaced term to
get
(25000 + .06*25000) ( 1 + .06)
Factor 25000 out to get
25000 * (1 + .06) * (1+ .06)
The last two terms multiple out to give (1 + .06)^2
so....
end of nth year salary is
25000 * ( 1 + .06) ^n
To double her salary we need to find "n" such that
50000 = 25000 ( 1.06) ^ n
Dividing both sides by 25000 gives
2 = 1.06 ^n
Taking the logarithm of both sides gives:
log2 = n * log (1.06) or
n = log2 / log(1/06)
This is approximately 0.3010 / 0.0253 or about 11.9 years.
Thank you,
MrB
