^{(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?

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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.

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

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