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aExponential funtion

Mathematics
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The Population size of a country was 12.7 million in the year 2000 and 14.3 million in the year 2010. Assume that the population size follows an exponential function. 

What will the population size be in the year 2015, assuming the formula holds until then?

When will the population reach 18 million?

Apr 16th, 2015

In t years after 2000 the population will be P(t) = P0ekt. Given P(0) = 12.7×106 and P(10) = 14.3×106, determine P0 = 12.7×106 and 14.3×106 = 12.7×106e10k. From the latter equation e10k = 14.3/12.7 and

k = ln(14.3/12.7)/10 = 0.01187. The population grows according to the exponential law as

P(t) = 12.7×106 e0.01187t.

In 2015 the population will be P(15) = 12.7×106 e0.01187×15= 15.2 million.

If P(t) = 18×106, then from the equation 18×106 = 12.7×106 e0.01187twe get e0.01187t = 18/12.7 = 1.417. Take the natural logarithm of both sides: 0.01187 t = ln(1.417) and t = ln(1.417)/0.01187 = 29.4 years

The population will reach 18 million in 2030.


Apr 15th, 2015

what is 0ekt?

Apr 15th, 2015

e^(kt) is the exponential function with the base e = 2.718... Do you mean that it is multiplied by zero?

Apr 15th, 2015

I just dont understand were the variables came from

Apr 15th, 2015

The independent variable is the time (in years) that passed since 2000 and the dependent variable is the population size. The population size is changing according to the exponential law, that is, P(t) = P_0 times e^(kt), where P_0 and k are some constants that can be found if two corresponding values of t and P(t) are given. If t = 0, then e^0 = 1, so P(0) = P_0 is the initial population (in 2000). 

The second pair t = 10 and P(10) = 14,300,000 was used to find the constant k. After both P_0 and k are found, it is possible to use the formula P(t) = P_0 times e^(kt) either to find P(t) for given t or to find t for given value of P(t).

Apr 15th, 2015

​(2/x-5) < (5/x+1) less than is actually less than or equal to

can you please help with this?

Apr 15th, 2015

Solve the inequality 2/x - 5 <= 5/x + 1          

Rewrite with the same denominator: (2 - 5x)/x <= (5 + x)/x. Subtract (2 - 5x)/x from both sides 

0 <= (5 + x)/x - (2 - 5x)/x = (5 + x - 2 + 5x)/x = (3 + 6x) /x

or 3(2x + 1)/x >= 0. The expression on the left-hand side changes it sign at the points x = - 1/2 (where 2x + 1 = 0) and x = 0. They divide the number line into three parts

x < -1/2 there both 2x + 1 and x are negative and their quotient is positive (note that at x = -1/2 the inequality holds, 0 >= 0)

-1/2 < x < 0  there 2x + 1 > 0 and x < 0, so the quotient (2x + 1)/x < 0

x > 0 both 2x + 1 and x are negative and their quotient is positive

Answer: x <= -1/2 and x > 0.

Apr 16th, 2015

What would be the domain?

Apr 16th, 2015

The solution of the inequality 2/x - 5 <= 5/x + 1 must be within the domain {x | x not 0} because both sides are undefined if x = 0. 

Apr 16th, 2015

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Apr 16th, 2015
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Apr 16th, 2015
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