Alexandra invested $20,000 in an account paying an interest rate of 6.2% compounded annually. 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
$27,200

5.1

Step-by-step explanation:

Compounded Annually:

A=P(1+r)^t

A=P(1+r)  

t

A=27200\hspace{35px}P=20000\hspace{35px}r=0.062

A=27200P=20000r=0.062

Given values

27200=

27200=

\,\,20000(1+0.062)^{t}

20000(1+0.062)  

t

Plug in values

27200=

27200=

\,\,20000(1.062)^{t}

20000(1.062)  

t

Add

\frac{27200}{20000}=

20000

27200

=

\,\,\frac{20000(1.062)^{t}}{20000}

20000

20000(1.062)  

t

Divide by 20000

1.36=

1.36=

\,\,1.062^t

1.062  

t

\log\left(1.36\right)=

log(1.36)=

\,\,\log\left(1.062^t\right)

log(1.062  

t

)

Take the log of both sides

\log\left(1.36\right)=

log(1.36)=

\,\,t\log\left(1.062\right)

tlog(1.062)

Bring exponent to the front

\frac{\log\left(1.36\right)}{\log\left(1.062\right)}=

log(1.062)

log(1.36)

=

\,\,\frac{t\log\left(1.062\right)}{\log\left(1.062\right)}

log(1.062)

tlog(1.062)

Divide both sides by log(1.062)

5.1116317=

5.1116317=

\,\,t

t

Use calculator

t\approx

t≈

\,\,5.1

5.1