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Question

Here is a word problem...

The population of Brownsville has grown according to the mathematical model y=720,500(1.022)^x....Where x is the number of years


If this trend continues ,,,using the model, in how many years the population of Brownsville reach 1,548,8000...

Please help ,,Thanks.

Answer 1

You have this equation:

y=720,500(1.022)^x

You want to plug in a population of 1,548,800 for y and solve for x.

1,548,800 = 720,500(1.022)^x

Begin by dividing both sides by 720,500:
1,548,800 / 720,500 = (1.022)^x

2.14961832 = 1.022^x

Now take the log() of both sides:
log(2.14961832) = log(1.022^x)

Using this rule: log(a^x) = x log(a) we can write:
log(2.14961832) = x * log(1.022)

Now divide both sides by log(1.022)
x = log(2.14961832) / log(1.022)

Plug this into your calculator and you get
x ≈ 35.17 years

P.S. I assume you meant 1,548,800 (I ignored the extra zero).

Answer 2

Hey there!

Here's the answer.

y=720500(1.022)^x --> Write the problem.
15488000=720500(1.022)^x --> Substitute 15488000 for y.
2816/131=1.022^x --> Divide 720500 on both sides of the equation and reduce the left side of the fraction.
log(2816/131)=log(1.022^x) --> Take the common log on both sides of the equation.
log(2816/131)=xlog(1.022) --> Use the property log(m^n)=n*log(m).
log(2816/131)/log(1.022)=x --> Divide both sides of the equation by log(1.022).
140.97=x --> Approximate the left side of the equation.
x=140.97 Switch the terms around.

So the answer is about 140.97 years or 141 years.

Hope it helps!

Answer 3

dude plug it in it ...confusing with the 1,588,0000... did you add an extra 0

154880000=720,500(1.022)^x
154880000/720,500 = 1.022^x
ln 154880000/720,500 = x ln 1.022
x= (ln 154880000/720,500)/( ln 1.022)
x=246.787yrs

Answer 4

15488000 = 720500(1.022)^x
21.49 = 1.022^x
log 21.49 = xlog1.022
x = log 21.49/log 1.022 = 141

Answer 5

Log 720,500 + x log 1.022 = log 1548800

x = 35.2 years

Answer 6

1,548,8000 = 720,500(1.022)^x
x = ln(1,548,8000 /720,500) /ln1.022, years