The population of a town is 20,000 people in the year 2000. how many people with live in the town in 2016 if the population increases at a rate of 6% every 2 years? round your answer to the nearest whole number.

31877 people with live in the town in 2016 if the population increases at a rate of 6% every 2 years

Step-by-step explanation:

The formula used will be

A(t) = P(1+r)^t

A(t) = Future value

P = population

r = rate

t = time

P= 20,000

r =6% or 0.06

t = 16

Since population is increased every 2 years, so t = 16/2 = 8

Putting value:

A(16) = 20,000(1+0.06)^8

A(16)= 20,000(1.06)^8

A(16) = 31876.9 ≈ 31877

So, 31877 people with live in the town in 2016 if the population increases at a rate of 6% every 2 years

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the solution of the equation is 2.23 ⇒ answer b

step-by-step explanation:

* lets revise the meaning of exponential function

- the form of the exponential function is y = ab^x, where a ≠ 0, b > 0 ,  

  b ≠ 1, and x is any real number

- it has a constant base b

- it has a variable exponent x

- to solve an exponential equation, take the log or ln of both sides,  

  and solve for the variable

* lets solve the problem

∵

- the base is 7 and the exponent is x

- insert ㏑ in both sides

∴

- use the rule

∴ x ㏑(7) = ㏑(77)

- to find x divide both sides by ㏑(7)

∴

* the solution of the equation is 2.23 to the nearest 2 decimal places