\(A=P(1+ \frac{r}{n})^ \frac{t}{n}\)
A=future amount
P=present amount
r=rate in decimal
n=number of times per year compounded
t=time in years
how many years to double
basically
A=2P at t=?
so we can simplify and ignore the pricipal given and do
\(2=(1+ \frac{r}{n})^ \frac{t}{n}\)
r=7%=0.07
n=2 (semiannualy means 2 times per year)
t=t
\(2=(1+ \frac{0.07}{2})^ \frac{t}{2}\)
\(2=(1+ 0.035)^ \frac{t}{2}\)
\(2=(1.035)^ \frac{t}{2}\)
take the ln of both sides
\(ln2=(\frac{t}{2})ln1.035\)
divide both sides by ln1.035
\(\frac{ln2}{ln1.035} = \frac{t}{2}\)
times 2 both sides
\(\frac{2ln2}{ln1.035} =t\)
use calculator
40.2976=t
nearest tenth
40.3
about 40.3 years
