Question

Suppose the population of trout in a certain stretch of a river is 4000. In about how many years will the population of trout be 700 if the decay rate is 35%? Use the equation 700 = (4000)(0.65)x and round the value of x to the nearest year.

Answer 1

4.97991230017 years is the final answer. I would suggest rounding your answer to the nearest tenth, so 5 years should be your correct answer.

Answer 2

Answer: 4 years

Step-by-step explanation:

The exponential decay equation is given by :_

$y=A(1-r)^x$, where A is the initial population and r is the rate of decay in x years .

The initial population of trout in a certain stretch of a river =4000

The rate of decay = 35% =0.35

To find the number of years (x) for the population of trout be 700, we put all the values of the above equation, we get

$700=4000(1-0.35)^x\\\\\Rightarrow700=4000(0.65)^x\\\\\Rightarrow (0.65)^x=\frac{700}{4000}\\\\\Rightarrow(0.65)^x=0.175\\\\\text{Taking log on both sides, we get}\\\\\Rightarrow x\log(0.65)=\log(0.175)\\\\\Rightarrow x(-0.187086643357)=-0.756961951314\\\\\Rightarrow x=\frac{-0.756961951314}{-0.187086643357}\\\\\Rightarrow\ x=4.0460502029\approx4\text{ years}$