question 12: approximately how long must one wait (to the nearest year) for an initial investment of $1,000…

question 12: approximately how long must one wait (to the nearest year) for an initial investment of $1,000 to triple in value (i.e., increase to $3000) if the investment earns 8% compounded annually? a) 9 years b) 14 years c) 22 years d) 25 years e) none of these are correct question 13 1.5 points 1.5 points

question 12: approximately how long must one wait (to the nearest year) for an initial investment of $1,000 to triple in value (i.e., increase to $3000) if the investment earns 8% compounded annually? a) 9 years b) 14 years c) 22 years d) 25 years e) none of these are correct question 13 1.5 points 1.5 points

Answer

Explanation:

Step1: Recall compound - interest formula

The compound - interest formula is $A = P(1 + r)^t$, where $A$ is the final amount, $P$ is the principal amount, $r$ is the annual interest rate (as a decimal), and $t$ is the number of years. Here, $P=$1000$, $A = 3\times1000=$3000$ (since the investment triples), and $r=0.08$. Substituting these values into the formula gives $3000 = 1000(1 + 0.08)^t$.

Step2: Simplify the equation

Divide both sides of the equation $3000 = 1000(1 + 0.08)^t$ by $1000$. We get $3=(1.08)^t$.

Step3: Take the natural logarithm of both sides

$\ln(3)=\ln(1.08^t)$. Using the property of logarithms $\ln(a^b)=b\ln(a)$, we can rewrite the right - hand side as $t\ln(1.08)$. So, $\ln(3)=t\ln(1.08)$.

Step4: Solve for $t$

$t=\frac{\ln(3)}{\ln(1.08)}$. We know that $\ln(3)\approx1.0986$ and $\ln(1.08)\approx0.07696$. Then $t=\frac{1.0986}{0.07696}\approx14.27$. Rounding to the nearest year, $t = 14$ years.

Answer:

B. 14 years