Kylie invested $290 in an account paying an interest rate of 2.5% 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 $390?

t= 11.9

Step-by-step explanation:

Compounded Continuously:

A=Pe^{rt}

A=Pe  

rt

A=390\hspace{35px}P=290\hspace{35px}r=0.025

A=390P=290r=0.025

Given values

390=

390=

\,\,290e^{0.025t}

290e  

0.025t

Plug in

\frac{390}{290}=

290

390

​  

=

\,\,\frac{290e^{0.025t}}{290}

290

290e  

0.025t

​  

Divide by 290

1.3448276=

1.3448276=

\,\,e^{0.025t}

e  

0.025t

\ln\left(1.3448276\right)=

ln(1.3448276)=

\,\,\ln\left(e^{0.025t}\right)

ln(e  

0.025t

)

Take the natural log of both sides

\ln\left(1.3448276\right)=

ln(1.3448276)=

\,\,0.025t

0.025t

ln cancels the e

\frac{\ln\left(1.3448276\right)}{0.025}=

0.025

ln(1.3448276)

​  

=

\,\,\frac{0.025t}{0.025}

0.025

0.025t

​  

Divide by 0.025

11.8506326=

11.8506326=

\,\,t

t

t\approx

t≈

\,\,11.9

11.9

Round to the nearest tenth of a year

step-by-step explanation:

we are given

since, we have to isolate x on anyone side

so, we can move all x terms on one side

subtract both sides by 2x

add both sides by 9

divide both sides by 4

we get

a.)the line for function b is steeper