Owen invested $63,000 in an account paying an interest rate of 6.5% compounded continuously.
Assuming no deposits or withdrawals are made,
how long would it take, to the nearest tenth of a
year,
for the value of the account to reach
$159,600?

14.3

Step-by-step explanation:

A=Pe^{rt}

A=Pe  

rt

A=159600\hspace{35px}P=63000\hspace{35px}r=0.065

A=159600P=63000r=0.065

Given values

159600=

159600=

\,\,63000e^{0.065t}

63000e  

0.065t

Plug in

\frac{159600}{63000}=

63000

159600

=

\,\,\frac{63000e^{0.065t}}{63000}

63000

63000e  

0.065t

Divide by 63000

2.5333333=

2.5333333=

\,\,e^{0.065t}

e  

0.065t

\ln\left(2.5333333\right)=

ln(2.5333333)=

\,\,\ln\left(e^{0.065t}\right)

ln(e  

0.065t

)

Take the natural log of both sides

\ln\left(2.5333333\right)=

ln(2.5333333)=

\,\,0.065t

0.065t

ln cancels the e

\frac{\ln\left(2.5333333\right)}{0.065}=

0.065

ln(2.5333333)

=

\,\,\frac{0.065t}{0.065}

0.065

0.065t

Divide by 0.065

14.3005532=

14.3005532=

\,\,t

t

t\approx

t≈

\,\,14.3

14.3

right

step-by-step explanation:

the probability is 0