(a) ok
(b) 160,500(.76)^10 = 10,318
how did you get 121980?
For (c), so far so good
160,500(.76)^x = 100
0.76^x = 1/1605
x log0.76 = log(1/1605) = -log1605
x = -log1605 / log0.76 = 26.89
Would someone please check my answers and then tell me how to do the last question?
The population of the city of Pearville, y, is decreasing according to the mathematical model y =160,500(.76) ^ x, where x is the number of years. (Round all answers to the nearest whole number)
a) What is the starting population of Pearville?
My answer would be 160,500 people
b) What would the population be in 10 years, if this trend continues?
My answer is y = 160,500(.76)^10 = 121,980 people
c) Use this model to predict about when the population of Pearville will be about 100 people.
I know I need to set y equal to 100 since y is representing the number of people, but I'm not sure how to solve for x?? As in, what are the steps to solve for x?
100 = 160,500(.76)^x
First divide both sides by 160,500?
I'm quite confused because x is an exponent..........
4 answers
For b, I put it into my calculator and it came out with that number - obviously not correct, so thank you.
For c, I don't know how to do that log stuff - is there another way?
For c, I don't know how to do that log stuff - is there another way?
no other way. That is how you solve exponential equations.
just as + and - are inverse operations, and * and /, so are
logs and exponents.
to solve exponents, take the log
to solve log equations, take exponents.
log(x+3) = 1
x+3 = 10^1
x = 7
check: log(7+3) = log10 = 1
just as + and - are inverse operations, and * and /, so are
logs and exponents.
to solve exponents, take the log
to solve log equations, take exponents.
log(x+3) = 1
x+3 = 10^1
x = 7
check: log(7+3) = log10 = 1
b. y = 160,500(0.76)^10 = 10318.
c. 100 = 160500(0.76)^x
x = 27
c. 100 = 160500(0.76)^x
x = 27
1 answer