Exponential Logarithmic
2^3=8 log_2 8=3
5^2=25 log_5 25=2
10^4=10,000 log_10 10,000=4
log_1,000,000=x log_1,000,000=x
e^x=12 ln 12=x
The population of a city (in thousands) closely fits the exponential function P(x)=500e^0.04x where x is the number of years since the year 2000.
Compute the expected population of the city in the year 2005:
To compute the expected population of the city in 2005, we substitute x = 5 (since it is the number of years since 2000) into the exponential equation P(x)=500e^0.04x.
P(5)=500e^0.04(5)
P(5)=500e^0.2
P(5)≈500(1.22140)
P(5)≈610.70
Therefore, the expected population of the city in the year 2005 is approximately 610,700.
After how many years would the population of the city reach 800 thousand:
To find the number of years it takes for the population to reach 800,000, we need to set up and solve the exponential equation P(x)=500e^0.04x=800.
500e^0.04x=800
e^0.04x=800/500
e^0.04x=1.6
Take the natural logarithm (ln) of both sides to solve for x:
ln(e^0.04x)=ln(1.6)
0.04x=ln(1.6)
x=(ln(1.6))/(0.04)
Using a calculator, we find that x is approximately 21.98.
Therefore, it would take approximately 22 years for the population of the city to reach 800,000.
Complete the table below comparing exponential equations with the equivalent logarithmic equation. Then first row is done for you.
b^x=y is the same as 〖log〗_b y=x
Exponential Logarithmic
2^3=8 log_2 8=3
5^2=25
10^4=10,000
log1,000,000=x
e^x=12
ln〖4=y〗
The population of a city (in thousands) closely fits the exponential function P(x)=500e^0.04x where x is the number of years since the year 2000.
Compute the expected population of the city in the year 2005.
After how many years would the population of the city reach 800 thousand?
1 answer
1 answer