5400 dollars is placed in an account with an annual interest rate of 9%. to the nearest tenth of a year, how…

Explanation:

Step1: Use 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 (in decimal form), and $t$ is the number of years. Here, $P = 5400$, $r=0.09$, and $A = 15500$. So we have the equation $15500=5400(1 + 0.09)^t$.

Step2: Simplify the equation

First, divide both sides of the equation by $5400$: $\frac{15500}{5400}=(1.09)^t$, which simplifies to $\frac{155}{54}=(1.09)^t$.

Step3: Take the natural logarithm of both sides

$\ln(\frac{155}{54})=\ln(1.09^t)$. Using the property of logarithms $\ln(a^b)=b\ln(a)$, we get $\ln(\frac{155}{54}) = t\ln(1.09)$.

Step4: Solve for $t$

$t=\frac{\ln(\frac{155}{54})}{\ln(1.09)}$. Calculate $\ln(\frac{155}{54})\approx\ln(2.87037)\approx1.053$ and $\ln(1.09)\approx0.0862$. Then $t=\frac{1.053}{0.0862}\approx12.2$.

Answer:

$12.2$ years