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Diano4ka-milaya [45]
2 years ago
12

HELP 100 POINTS The population of a city is modeled with the function P=250,000e^0.013t​, where t is the number of years since 2

000. In what year will the population reach ​450,000?
Mathematics
1 answer:
erma4kov [3.2K]2 years ago
6 0
Pretty difficult problem, but that’s why I’m here.
First we need to identify what we’re looking for, which is t. So now plug 450k into equation and solve for t.
450000 = 250000e^0.013t
Now to solve this, we need to remember this rule: if you take natural log of e you can remove x from exponent. And natural log of e is 1.
Basically ln(e^x) = xln(e) = 1*x
So knowing this first we need to isolate e
450000/250000 = e^0.013t
1.8 = e^0.013t
Now take natural log of both
Ln(1.8) = ln(e^0.013t)
Ln(1.8) = 0.013t*ln(e)
Ln(1.8) = 0.013t * 1
Now solve for t
Ln(1.8)/0.013 = t
T= 45.21435 years
Now just to check our work plug that into original equation which we get:
449999.94 which is basically 500k (just with an error caused by lack of decimals)
So our final solution will be in the 45th year after about 2 and a half months it will reach 450k people.
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\implies \sf \dfrac{17474}{7790}=\left(1.0125}\right)^{12t}

\implies \sf \ln\left(\dfrac{17474}{7790}\right)=\ln \left(1.0125}\right)^{12t}

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\implies \sf t=\dfrac{\ln\left(\frac{17474}{7790}\right)}{12 \ln \left(1.0125}\right)}

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