SOLUTION: The population P of a fish farm in t years is modeled by the equation P(t) = 2000/1 + 9e exponent is −0.6t . To the nearest tenth, how long will it take for the population

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Question 1153619: The population P of a fish farm in t years is modeled by the equation
P(t) =
2000/1 + 9e exponent is −0.6t
.
To the nearest tenth, how long will it take for the population to reach 900?
Found 3 solutions by jim_thompson5910, MathLover1, MathTherapy:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

Plug in P(t) = 900 and solve for t
P(t) = 2000/(1 + 9*e^(-0.6*t))
900 = 2000/(1 + 9*e^(-0.6*t))
900(1 + 9*e^(-0.6*t)) = 2000
900*1 + 900*9*e^(-0.6*t) = 2000
900 + 8100*e^(-0.6*t) = 2000
8100*e^(-0.6*t) = 2000-900
8100*e^(-0.6*t) = 1100
e^(-0.6*t) = (1100)/(8100)
e^(-0.6*t) = 0.135802469135802
Ln( e^(-0.6*t) ) = Ln( 0.135802469135802 )
-0.6*t*Ln( e ) = Ln( 0.135802469135802 )
-0.6*t*1= -1.99655388187407
-0.6*t= -1.99655388187407
t= (-1.99655388187407)/(-0.6)
t= 3.32758980312345


If we plug t= 3.32758980312345 into the function, we get
P(t) = 2000/(1+9*e^(-0.6*t))
P(3.32758980312345) = 2000/(1+9*e^(-0.6*3.32758980312345))
P(3.32758980312345) = 2000/(1+9*e^(-1.99655388187407))
P(3.32758980312345) = 2000/(1+9*0.135802469135802)
P(3.32758980312345) = 2000/(1+1.22222222222222)
P(3.32758980312345) = 2000/(2.22222222222222)
P(3.32758980312345) = 900.000000000001
which is really really close to 900. The error is due to rounding

When rounding 3.32758980312345 to the nearest tenth, you should get 3.3

So the answer is approximately 3.3 years

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
The population P of a fish farm in t years is modeled by the equation:
P%28t%29+=2000%2F%281+%2B+9e%5E%28-0.6t%29%29
To the nearest tenth, how long will it take for the population to reach 900?

find P%28t%29+=900

900+=2000%2F%281+%2B+9e%5E%28-0.6t%29%29........multiply both sides by %281+%2B+9e%5E%28-0.6t%29%29

900%2A%281%2B9e%5E%28-0.6t%29%29+=2000......divide both sides by 900

1%2B9e%5E%28-0.6t%29=20%2F9...move 1 to the right side

9e%5E%28-0.6t%29=20%2F9-1

9e%5E%28-0.6t%29=20%2F9-9%2F9

9e%5E%28-0.6t%29=11%2F9

e%5E%28-0.6t%29=%2811%2F9%29%2F9

e%5E%28-0.6t%29=11%2F81....take natural logarithm of both sides

ln%28e%5E%28-0.6t%29%29=ln%2811%2F81%29

%28-0.6t%29+ln%28e%29=ln%2811%29-ln%2881%29.....ln%28e%29=1

-0.6t=ln%2811%29-ln%2881%29

t=%28ln%2811%29-ln%2881%29%29%2F-0.6

t ≈ 3.32759+

To the nearest tenth, it will take 3.3 for the population to reach 900


Answer by MathTherapy(10556) About Me  (Show Source):
You can put this solution on YOUR website!
The population P of a fish farm in t years is modeled by the equation
P(t) =
2000/1 + 9e exponent is −0.6t
.
To the nearest tenth, how long will it take for the population to reach 900?
matrix%281%2C3%2C+P%28t%29%2C+%22=%22%2C+%222%2C000%22%2F%281+%2B+9e%5E%28-+.6t%29%29%29

 ----- Substituting 900 for P(t), and reducing the fractions

matrix%281%2C3%2C+9%281+%2B+9e%5E%28-+.6t%29%29%2C+%22=%22%2C+20%29 ---- Cross-multiplying

matrix%281%2C3%2C+1+%2B+9e%5E%28-+.6t%29%2C+%22=%22%2C+20%2F9%29 -------- Dividing each side by 9

matrix%281%2C3%2C+9e%5E%28-+.6t%29%2C+%22=%22%2C+20%2F9+-+1%29 -------- Subtracting 1 from each side

matrix%281%2C3%2C+9e%5E%28-+.6t%29%2C+%22=%22%2C+11%2F9%29

matrix%281%2C3%2C+e%5E%28-+.6t%29%2C+%22=%22%2C+%2811%2F9%29%2F9%29 ---------- Dividing each side by 9

matrix%281%2C3%2C+e%5E%28-+.6t%29%2C+%22=%22%2C+11%2F81%29

matrix%281%2C3%2C+-+.6t%2C+%22=%22%2C+ln+%2811%2F81%29%29 ---------- Converting to LOGARITHMIC (Natural) form