5500 dollars is placed in an account with an annual interest rate of 6.5%. To the nearest tenth of a year, how long will it take for the account value to reach 19700 dollars?

Accepted Solution

Answer:t = 20.3 yearsStep-by-step explanation:I am assuming that this amount of money invested is compounding annually, so I am going to use the formula that goes along with that assumption:[tex]A(t)=P(1+r)^t[/tex]where A(t) is the amount at the end of compounding, P is the initial investment, r is the interest rate in decimal form, and t is the time in years.  We are solving for t.  Right now it is the exponent, but we have to get it down from that position in order to solve for it.  The only way we can do that is to eventually take the natural log of both sides.  But let's write the equation first and then do some simplifying to make things a bit easier mathematically:[tex]19,700=5,500(1+.065)^t[/tex] and[tex]19,700=5,500(1.065)^t[/tex]We will divide both sides by 5,500:[tex]3.58181818=(1.065)^t[/tex]Taking the natural log of both sides gives us:[tex]ln(3.58181818)=ln(1.065)^t[/tex]The power rule for logs (both common and natural) tells us that once we take the log or ln of a base, the exponent comes down out front:ln(3.58181818) = t ln(1.065)Now we can divide both sides by ln(1.065) and do the math on our calculators to find that t = 20.2600 or, to the tenth of a year, t = 20.3 years