Question

A person invests $200 in an account that earns 1.98% annual interest compounded
quarterly. Find when the value of the investment reaches $500.

Answer 1

Answer:

$t=46.4\ years$

Step-by-step explanation:

we know that    

The compound interest formula is equal to  

$A=P(1+\frac{r}{n})^{nt}$  

where  

A is the Final Investment Value  

P is the Principal amount of money to be invested  

r is the rate of interest  in decimal

t is Number of Time Periods  

n is the number of times interest is compounded per year

in this problem we have  

$t=?\ years\\P=\ 200\\A=\ 500\\ r=1.98\%=1.98/100=0.0198\\n=4$  

substitute in the formula above

$500=200(1+\frac{0.0198}{4})^{4t}$  

$2.5=(1.00495)^{4t}$  

Apply property of exponents

$2.5=[(1.00495)^{4}]^t$  

Apply log both sides

$log(2.5)=log[(1.00495)^{4}]^t$  

$t=log(2.5)/log[(1.00495)^{4}]$  

$t=46.4\ years$