If P dollars are invested at the end of each year in an annuity that earns interest at an annual rate r, the amount in the account will be A dollars after n years, where $$n = \frac { \log \left[ \frac { A r } { p } + 1 \right] } { \log ( 1 + r ) }$$ If $2,600 is invested each year in an annuity earning 14% annual interest, when will the account be
Worth $30,000?
A)2.7 years
B)7.3 years
C)11.5 years
D)3.7 years
E)16.9 years

Question

Quizplus · Accepted Answer

Using the given formula, we substitute $P = 2600$, $A = 30000$, and $r = 0.14$ (14% interest rate converted to decimal form) into the equation to solve for $n$. The calculation will yield a result close to 7.3 years, making option B the correct answer.

If P Dollars Are Invested at the End of Each n=log⁡[Arp+1]log⁡(1+r)n = \frac { \log \left[ \frac { A r } { p } + 1 \right] } { \log ( 1 + r ) }n=log(1+r)log[pAr​+1]​

If P Dollars Are Invested at the End of Each $n = \frac { \log \left[ \frac { A r } { p } + 1 \right] } { \log ( 1 + r ) }$