The U.S. population in 1990 was approximately 250 million, and the average growth

rate for the past 30 years gives a doubling time of 66 years. The above formula for the
United States then becomes
P (in millions)= 250 x 2( y-1990)/66

1.What was the approximate population of the United States in 1960?

2. What will the population of the United States be in 2025 if this
growth rate continues?

P (in millions)= 250 x 2^[( y-1990)/66]

Insert y = 1960 and y = 2025 in here.

yes i do that but I do not get a correct answer. 2025-1990=35/66=.5303
1960-1990=-30/36=-.4545
what do i do from there I know the answer for the 1960 in 182 million but I am still stuck on how they got that answer

250*2^(0.5303) = ...? (use your calculator)

And:

250 million *2^(-0.4545) = 182 million

So it is the square root -0.4545

i appreciate your help

No, it's 2 to the power -0.4545.

I think you know that:

a^n = a*a*a*... *a (n factors of a)

But here n must be a positive integer. It turns out one can "continue" the power function allowing n to take on any real value and not just a positive integer.

I know I am asking a lot of question but let say -0.4545, how exactly do you figure that out, I do not have a calucator that I can push in the numbers, I use the one on my computer, that is why I am asking.

Thank You Again For your Help

If you have the e^x (or Exp) function and the ln(x) function, then you can use that:

2^(x) = e^[x*ln(2)] Or

2^(x) = Exp[x*ln(2)]

If you have just addition multiplication and division, then you can still compute it using series expansions. I can explain that later, if necessary...

If you don't have a calculator that can directly compute exponential functions, you can use the series expansion of the exponential function to approximate the value of 2^(-0.4545):

2^(-0.4545) ≈ 1 - 0.4545 * ln(2) + (0.4545^2 * ln(2)^2) / 2

You can use the value of ln(2) as approximately 0.6931:

2^(-0.4545) ≈ 1 - 0.4545 * 0.6931 + (0.4545^2 * 0.6931^2) / 2

Calculating this expression will give you an approximate value for 2^(-0.4545), which you can then use to determine the population of the United States in 1960 or any other year you're interested in.

No problem at all! If you don't have access to a calculator or specific functions like e^x or ln(x) on your computer, you can still approximate the value of 2^(-0.4545) using a method called the power series expansion.

To do this, you can use the following formula:

1 + x + x^2/2! + x^3/3! + x^4/4! + ...

In this case, your x will be -0.4545. You can start by evaluating the terms of the series one by one, starting from the first term, which is just 1:

1 + (-0.4545) + (-0.4545)^2/2! + (-0.4545)^3/3! + ...

Calculating each term separately and summing them up will give you an approximation of 2^(-0.4545). The more terms you calculate, the more accurate your approximation will be.

However, if you have access to an online scientific calculator or a calculator application on your computer or smartphone, it would be much simpler and faster to use that to directly calculate 2^(-0.4545) and get the exact value.