^{(n-1)}

^{(n-1)}

An oil spill in the ocean is increasing in its surface area by 21% each day. It begins at two square miles. How long before it reaches 10 square miles?

Tutors, sign in to answer this question.

The oil spill grow will generate a series which is

A=Ao (1.21)^{(n-1)}

A=10

Ao=2

5=(1,21)^{(n-1)}

(n-1) = log5/log1,21

n-1 = 8,44

n= 7.44 days

Area of Oil Spill = 2 square miles when time (t) = 0

Area of Oil Spill after 1 day = 2 x (1.21)

STEP 1: Establish the formula

Area of Oil Spill after t days = 2 x (1.21)^t (raised to the power of t)

STEP 2: If the area of the oil spill is 10 square miles, set the formula equal to 10.

10 = 2 x (1.21)^t

STEP 3: Take the logarithm of both sides

log 10 = log [2 x(1.21)^t]

1 = log (2) + t x log (1.21)

REASON: 10^1 = 10

the log of two amounts that are multiplied together equals the sum of the logs.

EXAMPLE: log (a x b) = log a + log b

And the log of an amount raised to a power is equal to that power times the log

EXAMPLE: log b^2 = 2 x log b

STEP 4: Find the log values of the numbers

log(2) = .3010 log (1.21) = .0828

STEP 5: Plug the log values into the formula

1 = .3010 + t x (.0828)

STEP 6: Subtract .3010 from each side of the equation

.699 = t x (.0828)

Step 7: Divide each side by .0828

t = 8.442

Area of Oil Spill after 1 day = 2 x (1.21)

STEP 1: Establish the formula

Area of Oil Spill after t days = 2 x (1.21)^t (raised to the power of t)

STEP 2: If the area of the oil spill is 10 square miles, set the formula equal to 10.

10 = 2 x (1.21)^t

STEP 3: Take the logarithm of both sides

log 10 = log [2 x(1.21)^t]

1 = log (2) + t x log (1.21)

REASON: 10^1 = 10

the log of two amounts that are multiplied together equals the sum of the logs.

EXAMPLE: log (a x b) = log a + log b

And the log of an amount raised to a power is equal to that power times the log

EXAMPLE: log b^2 = 2 x log b

STEP 4: Find the log values of the numbers

log(2) = .3010 log (1.21) = .0828

STEP 5: Plug the log values into the formula

1 = .3010 + t x (.0828)

STEP 6: Subtract .3010 from each side of the equation

.699 = t x (.0828)

Step 7: Divide each side by .0828

t = 8.442

The answer is 8.442 days.

Philip P. | Effective and Affordable Math TutorEffective and Affordable Math Tutor

This is an exponential growth problem:

Area = (2 mi^{2})*(1.21)^{t}

Where t is the number of days after the spill. Solve for t when the Area = 10 mi^{2}:

10 = 2*(1.21)^{t}

5 = (1.21)^{t}

ln(5) = t*ln(1.21)

ln(5)/ln(1.21) = t

Already have an account? Log in

By signing up, I agree to Wyzant’s terms of use and privacy policy.

Or

To present the tutors that are the best fit for you, we’ll need your ZIP code.

Your Facebook email address is associated with a Wyzant tutor account. Please use a different email address to create a new student account.

Good news! It looks like you already have an account registered with the email address **you provided**.

It looks like this is your first time here. Welcome!

To present the tutors that are the best fit for you, we’ll need your ZIP code.

Please try again, our system had a problem processing your request.

## Comments

0 2.00

1 2.42

2 2.93

3 3.54

4 4.29

5 5.19

6 6.28

7 7.59

8 9.19

9 11.12

10 13.45