Question

In how many years will the population of a city be 120,000 if the population P in thousands of cities can be modeled by the equation $P=80{{e}^{0.015t}}$ where t is the time in years?

Answer 1

Substitute the value of P equal to 120,000 in the relation $P=80{{e}^{0.015t}}$ and multiply the R.H.S with 1000. Now, take natural log, i.e. log to the base e, both the sides and use the property of logarithm $\ln {{e}^{m}}=m$ to find a linear relation in t. Solve for the value of t by using the value $\ln \left( \dfrac{3}{2} \right)=0.405$ for the calculations to get the answer.

Complete step-by-step solution:
Here we are provided with the relation of the population of a town that varies exponentially according to the relation $P=80{{e}^{0.015t}}$. We have to determine the value of time t when the population of the city will become 120,000.
Now, it is given that the population P is measured in thousands and time t is in years, so at time t the population relation $P=80{{e}^{0.015t}}$ means that the actual number of people is $P=80{{e}^{0.015t}}\times 1000$. So substituting 120,000 in place of P we get,
$\Rightarrow 120000=80{{e}^{0.015t}}\times 1000$
On simplifying we get,
$\Rightarrow \dfrac{3}{2}={{e}^{0.015t}}$
Taking natural log, i.e. log to the base e both the sides we get,
$\Rightarrow \ln \left( \dfrac{3}{2} \right)=\ln \left( {{e}^{0.015t}} \right)$
Using the formula $\ln {{e}^{m}}=m$ we get,
$\Rightarrow \ln \left( \dfrac{3}{2} \right)=0.015t$
$\Rightarrow t=\dfrac{\ln \left( \dfrac{3}{2} \right)}{0.015}$
Substituting the value $\ln 1.5=0.405$ and simplifying we get,

& \Rightarrow t=\dfrac{0.405}{0.015} \\\ & \therefore t=27 \\\ \end{aligned}$$ **Hence, after t = 27 years the population of the town will be 120,000.** **Note:** You may make a mistake in understanding the words of the question and directly substituting $120000=80{{e}^{0.015}}t$ but clearly it is said that the population P is in thousand so the correct relation will be $$120000=80{{e}^{0.015t}}\times 1000$$. Remember that we have used a log table to substitute the approx value of $\ln \left( \dfrac{3}{2} \right)$. You can also use the relation $\ln x=2.303{{\log }_{10}}x$ and the approximate values $\log 3=0.47,\log 2=0.30$ to get the answer. There may be some decimal differences which can be rounded off to get the answer. Remember the formula $\log \left( \dfrac{m}{n} \right)=\log m-\log n$.