Jordan invested $97,000 in an account paying an interest rate of 1.6% compounded monthly. Assuming no deposits or withdrawals are made, how long would it take, to
the nearest year, for the value of the account to reach $109,700?

t≈ 8

Step-by-step explanation:

\text{Compounded Monthly:}

Compounded Monthly:

A=P\left(1+\frac{r}{n}\right)^{nt}

A=P(1+  

n

r

)  

nt

Compound interest formula

A=109700\hspace{35px}P=97000\hspace{35px}r=0.016\hspace{35px}n=12

A=109700P=97000r=0.016n=12

Given values

109700=

109700=

\,\,97000\left(1+\frac{0.016}{12}\right)^{12t}

97000(1+  

12

0.016

)  

12t

Plug in values

109700=

109700=

\,\,97000(1.0013333)^{12t}

97000(1.0013333)  

12t

Simplify

\frac{109700}{97000}=

97000

109700

=

\,\,\frac{97000(1.0013333)^{12t}}{97000}

97000

97000(1.0013333)  

12t

Divide by 97000

1.1309278=

1.1309278=

\,\,1.0013333^{12t}

1.0013333  

12t

\log\left(1.1309278\right)=

log(1.1309278)=

\,\,\log\left(1.0013333^{\color{blue}{12t}}\right)

log(1.0013333  

12t

)

Take the log of both sides

\log\left(1.1309278\right)=

log(1.1309278)=

\,\,\color{blue}{12t}\log\left(1.0013333\right)

12tlog(1.0013333)

Bring exponent to the front

\frac{\log\left(1.1309278\right)}{\log\left(1.0013333\right)}=

log(1.0013333)

log(1.1309278)

=

\,\,\frac{12t\log\left(1.0013333\right)}{\log\left(1.0013333\right)}

log(1.0013333)

12tlog(1.0013333)

Divide both sides by log(1.0013333)

92.3402971=

92.3402971=

\,\,12t

12t

Use calculator

\frac{92.3402971}{12}=

12

92.3402971

=

\,\,\frac{12t}{12}

12

12t

Divide by 12

7.6950248=

7.6950248=

\,\,t

t

554.4

step-by-step explanation:

44*12.6=544.4

no one knows lol

step-by-step explanation: