17. Solve 1n 2 + 1n x = 5. Round to the nearest thousandth, if necessary. (2 points)

18. The sales of lawn mowers t years after a particular model is introduced is given by the function y = 5500 ln(9t + 4), where y is the number of mowers sold. How many mowers will be sold 5 years after a model is introduced? Round the answer to the nearest whole number. (2 points)

19.

In a particular region of a national park, there are currently 330 deer, and the population is increasing at an annual rate of 11%.

a. Write an exponential function to model the deer population.
b. Explain what each value in the model represents.
c. Predict the number of deer that will be in the region after five years. Show your work. bentleah

Posts: 53
Joined: April 11th, 2012, 9:18 am

Solve ln 2 + ln x = 5. Round to the nearest thousandth

ln(2x) = 5

2x = e^5

x = (1/2)e^5, about 74.207
=======================================…
Evaluate y = 5500 ln(9t + 4) for t = 5

5500ln(45+4) = 5500ln(49), about 21,405
=======================================…
P(t) = 330*(1.11)^t

330 is the initial value, the 0.11 is the growth rate and t is time in years.

P(5) = 330*(1.11)^5 Use calculator: corbet

Posts: 96
Joined: October 28th, 2011, 3:44 am

OK I assume you meant ln(2) + ln(x) = 5
17. by properties of ln's the LHS = ln(2x) and that = 5
Now, let these be exponents of e and you have 2x = e^5 so x e^5/2 as your answer.

18. y = 5500(ln(9t+4) at 5 years you have 5500*(ln(49)) 5500*3.892 = 21405 mowers is your answer.

19. This can be represented by the compound interest formula
D(t) = D(0)*(1+(r/n)^(n*t) D(t) = number of deer at year t
D(0) = number of deer at year 0.
n is the number of compoundind periods in a year...just 1 in your case
r is the rate of increase but in decimal form.
t is time in years.
Therefore you have D(5) = 330*(1+0.11)^5 which gives you 556 deer.

D(5) = 330*e^(0.11*5) = 572 rounded up. aldrick

Posts: 57
Joined: March 15th, 2012, 6:32 pm