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

infinitely many solutions

step-by-step explanation:

we have been given the equation;

2(2x + 3) = -4 + 2(2x + 5)

we need to determine the value of x.

2(2x + 3), we expand this expression to obtain 4x + 6

similarly 2(2x + 5) will become 4x + 10

now we have;

4x + 6 = -4 + 4x + 10

4x + 6 = 4x + 10 - 4

4x + 6 = 4x + 6

the expression on the left hand side is identical to the one on the right hand side. therefore, we are looking at just a single straight line whose equation is;

y = 4x + 6

consequently, we shall have infinitely many solutions