Bridge to College Math Unit 6 Task #21

Jack is investing $5000 in an account that earns 4% interest compounded annually. Determine to the nearest month when the investment will be worth $8000. Which of the following would be the equation used to solve Jack's equations.

• 5000= .04(x) + 8000

• 8000= 1.04(x) + 5000

• 8000= 5000(.04)^x

• 8000= 5000(1.04)^x

• 5000=8000(.96)^x

Jack is investing $5000 in an account that earns 4% interest compounded annually. Determine to the nearest month when the investment will be worth $8000. Which Logarithms could be used to solve the following problem, Choose all that apply.

• $Log_{1.04}1.6=x$

• $Log_{1.04}\ x\ =1.6$

• $Log_{1.06}1.04=x$

• $\frac{Log\left(1.04\right)}{Log\ \left(1.6\right)}=x$

• $\frac{Log\ \left(1.6\right)}{Log\ \left(1.04\right)}=x$

Jack is investing $5000 in an account that earns 4% interest compounded annually. Determine to the nearest year when the investment will be worth $8000.

When Bev was born, her grandma invested $4000 in an account that earns 5% interest compounded annually. When will the account double in value?  Choose all that apply.

•  $2=1.05\left(x\right)+4000$  

•  $=1.05\left(x\right)+4000$  

•  $2=\left(1.05\right)^x$  

•  $8000=4000\left(5\right)^x$  

•  $4000=8000\left(1.05\right)^x$  

When Bev was born, her grandma invested $4000 in an account that earns 5% interest compounded annually. Which logarithms could you use to solve this problem?

•  $Log_{1.05}\ 2\ =x$  

•  $Log_{4000}8000=2$  

•  $\frac{Log\left(2\right)}{Log\ \left(1.05\right)}=x$  

•  $\frac{Log\left(5\right)}{Log\left(2\right)}\ =x$  

•  $\frac{Log\ .05}{Log\ 2}=x$  

When Bev was born, her grandma invested $4000 in an account that earns 5% interest compounded annually. When will the account double in value, please enter your answer to the nearest tenth of a year

You bought a wake board boat 4 years ago for $45,000. The boat’s value depreciates by 10% per year. (a) How much is it worth today? Which equation would be used to solve this problem?

• $f\left(x\right)=45000\left(.90\right)^4$

• $f\left(x\right)=45000\left(1.10\right)^4$

• $4=45000\left(1.10\right)^x$

• $45000=4\left(1.10\right)^x$

• $4=45000\left(.9\right)^x$

You bought a wake board boat 4 years ago for $45,000. The boat’s value depreciates by 10% per year. How long until it is worth $20,000? Which equation could be used to solve this problem?

• $20000=45000\left(.9\right)^x$

• $45000=20000\left(1.1\right)^x$

• $=-.9\left(x\right)+45000$

• $=1.1\left(x\right)+20000$

• $20000=45000\left(.1\right)^x$

You bought a wake board boat 4 years ago for $40,000. The boat’s value depreciates by 10% per year. How long until it is worth $20,000? Which logarithm would be used to solve this problem?

•  $\frac{Log\ \left(.5\right)}{Log\ \left(.9\right)}=x$ 

•  $Log_{.9}\left(.5\right)=x$ 

•  $Log_2\left(1.1\right)=x$ 

•  $\frac{Log\ \left(2\right)}{Log\ \left(.9\right)}=x$ 

•  $\frac{Log\left(1.1\right)}{Log\ \left(2\right)}$ 

You bought a wake board boat 4 years ago for $40,000. The boat’s value depreciates by 10% per year.  How many years to the nearest tenth of a year will it take until it is worth $20,000