[tex]A=P(1+ \frac{r}{n})^ \frac{t}{n} [/tex]
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
[tex]2=(1+ \frac{r}{n})^ \frac{t}{n} [/tex]
r=7%=0.07
n=2 (semiannualy means 2 times per year)
t=t
[tex]2=(1+ \frac{0.07}{2})^ \frac{t}{2} [/tex]
[tex]2=(1+ 0.035)^ \frac{t}{2} [/tex]
[tex]2=(1.035)^ \frac{t}{2} [/tex]
take the ln of both sides
[tex]ln2=(\frac{t}{2})ln1.035 [/tex]
divide both sides by ln1.035
[tex] \frac{ln2}{ln1.035} = \frac{t}{2} [/tex]
times 2 both sides
[tex] \frac{2ln2}{ln1.035} =t [/tex]
use calculator
40.2976=t
nearest tenth
40.3
about 40.3 years
