Question

1000 was invested at 5.5% interest, compounded annually. After sometime the amount had grown to 1550. How long was the money invested?

Answer 1

compound interest equation for annually compounded

$A=P(1+r)^t$

A=final amount

P=principal

r=rate in decimal

t=time in years

given that

A=1550

P=1000

r=5.5%=0.055

find t

$1550=1000(1+0.055)^t$

divide both sides by 1000

$1.55=1.055^t$

take ln of both sides

$ln(1.55)=ln(1.055^t)$

use ln rule $ln(a^b)=b(ln(a))$

$ln(1.55)=t(ln(1.055))$

divide both sides by ln(1.055)

$\frac{ln(1.55)}{ln(1.055)}=t$

using a calculator, we get that t=8.18544 yrs

so about 8.2yrs