MATH SOLVE

4 months ago

Q:
# The population of a city is growing according to the exponential model p = cekt, where p is the population in thousands and t is measured in years. if the population doubles every 11 years what is k, the city's growth rate? [round answer to the nearest hundredth.]a.2.8%b.4.4%c.6.3%d.8.9%

Accepted Solution

A:

p = ce^(kt)

make p = 2c because doubling will be

like c-->2c, so if p = 2c, then

p = ce^(kt)

2c = ce^(kt)

2c/c = (ce^(kt))/c

2 = e^(kt)

Now take natural logarithm (ln) of both sides of the equation:

ln (2) = ln (e^(kt))

0.693 = kt×ln e

**this is because ln of an exponent makes the exponent become multiplied by the ln,

and ln e = 1

0.693 = kt×ln e

0.693 = kt×1, and t = 11 years

0.693 = k(11)

0.693/11 = 11k/11

k = 0.063, multiply by 100 to get %

k = 6.3%

answer is C

make p = 2c because doubling will be

like c-->2c, so if p = 2c, then

p = ce^(kt)

2c = ce^(kt)

2c/c = (ce^(kt))/c

2 = e^(kt)

Now take natural logarithm (ln) of both sides of the equation:

ln (2) = ln (e^(kt))

0.693 = kt×ln e

**this is because ln of an exponent makes the exponent become multiplied by the ln,

and ln e = 1

0.693 = kt×ln e

0.693 = kt×1, and t = 11 years

0.693 = k(11)

0.693/11 = 11k/11

k = 0.063, multiply by 100 to get %

k = 6.3%

answer is C