Question

jackie is an ornithologist, or a scientist who studies birds. every year, she measures the vulture population of the crystal plains. when she first measured the population, there were about 29,000 vultures living there. since then, she estimates that the vulture population is declining by about $\frac{1}{4}$ each year. write an exponential equation in the form $y = a(b)^x$ that can model the estimated vulture population, $y$, $x$ years after jackie started studying it. use whole numbers, decimals, or simplified fractions for the values of a and b. if this rate of decline continues, how many years after the first measurement will the vulture population measure below 15,000?

Explanation:

Step1: Identify the initial - value 'a'

The initial vulture population is 29000, so \(a = 29000\).

Step2: Identify the decay - factor 'b'

The population is declining by \(\frac{1}{4}\) each year. So the remaining fraction of the population each year is \(1-\frac{1}{4}=\frac{3}{4}= 0.75\), so \(b = 0.75\). The exponential equation is \(y = 29000(0.75)^x\).

Step3: Solve for the number of years 'x' when \(y\lt15000\)

Set up the inequality \(15000>29000(0.75)^x\). First, divide both sides by 29000: \(\frac{15000}{29000}>(0.75)^x\), which simplifies to \(\frac{15}{29}>(0.75)^x\).
Take the natural logarithm of both sides: \(\ln(\frac{15}{29})>x\ln(0.75)\).
Since \(\ln(0.75)<0\), when we divide both sides by \(\ln(0.75)\) the inequality sign flips. \(x>\frac{\ln(\frac{15}{29})}{\ln(0.75)}\).
\(\ln(\frac{15}{29})\approx\ln(15)-\ln(29)\approx2.708 - 3.367=- 0.659\), and \(\ln(0.75)\approx - 0.288\).
\(x>\frac{-0.659}{-0.288}\approx2.29\). Since \(x\) represents the number of years and it must be an integer, \(x = 3\).

Answer:

\(y = 29000(0.75)^x\)
3