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A person invested $5,500 in an account growing at a rate allowing the money to double every 14 years. How long, to the nearest tenth of a year would it take for the value of the account to reach $6,800?

Respuesta :

Answer:

t=4.3

Step-by-step explanation:

y=a(2)^{\frac{t}{d}}

y=a(2)  

d

t

​  

 

 

y=6800\hspace{40px}a=5500\hspace{40px}d=14

y=6800a=5500d=14

d is the doubling time

\text{Plug in:}

Plug in:

6800=

6800=

\,\,5500(2)^{\frac{t}{14}}

5500(2)  

14

t

​  

 

 

\text{Solve for }t\text{:}

Solve for t:

\frac{6800}{5500}=

5500

6800

​  

=

\,\,\frac{5500(2)^{\frac{t}{14}}}{5500}

5500

5500(2)  

14

t

​  

 

 

​  

 

Divide by 5500

1.23636364=

1.23636364=

\,\,2^{\frac{t}{14}}

2  

14

t

​  

 

 

\log(1.23636364)=

log(1.23636364)=

\,\,\log\left(2^{\color{green}{\frac{t}{14}}}\right)

log(2  

14

t

​  

 

)

Take the log of both sides

\log(1.23636364)=

log(1.23636364)=

\,\,\color{green}{\frac{t}{14}}\log(2)

14

t

​  

log(2)

Bring exponent to the front

\frac{\log(1.23636364)}{\log(2)}=

log(2)

log(1.23636364)

​  

=

\,\,\frac{t}{14}

14

t

​  

 

Divide by log(2)

0.30610313=

0.30610313=

\,\,\frac{t}{14}

14

t

​  

 

Divide in calculator

14(0.30610313)=

14(0.30610313)=

 

Multiply by 14

4.285443788=

4.285443788=

\,\,t

t

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