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3 Answers

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
 
 
 

Comments

The spill has an area of 2 sq mi at t=0.  After one day, t=1, it has increased by 21% to 2.42 sqmi.  So it should be 8.44 days, not 7.44.
 
day area
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
 
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
 
  The answer is 8.442 days.
This is an exponential growth problem:
 
Area = (2 mi2)*(1.21)t
 
Where t is the number of days after the spill.  Solve for t when the Area = 10 mi2:
 
10 = 2*(1.21)t
 
5 = (1.21)t
 
ln(5) = t*ln(1.21)
 
ln(5)/ln(1.21) = t
 
8.44 days = t