Question

A huge ice glacier in the Himalayas initially covered an area of 454545 square kilometers. Because of changing weather patterns, this glacier begins to melt, and the area it covers begins to decrease exponentially.
The relationship between AAA, the area of the glacier in square kilometers, and ttt, the number of years the glacier has been melting, is modeled by the following equation.
A=45e^{-0.05t}A=45e
−0.05t
A, equals, 45, e, start superscript, minus, 0, point, 05, t, end superscript
How many years will it take for the area of the glacier to decrease to 151515 square kilometers?
Give an exact answer expressed as a natural logarithm.

Answer 1

Answer:

$t=-20ln\left(\dfrac{1}{3}\right)$

Step-by-step explanation:

The relationship between A, the area of the glacier in square kilometers, and t, the number of years the glacier has been melting, is modeled by the equation.:

$A=45e^{-0.05t}$

We want to determine the value of t for which the area, A(t)=15 square kilometers.

$15=45e^{-0.05t}\\ Divide both sides by 45\\\dfrac{15}{45} =\dfrac{45e^{-0.05t}}{45}\\\dfrac{1}{3}=e^{-0.05t}\\ Take the natural logarithm of both sides\\ln\left(\dfrac{1}{3}\right)=ln\left(e^{-0.05t}\right)\\ln\left(\dfrac{1}{3}\right)=-0.05t\\ Divide both sides by -0.05 \\t=-\dfrac{ln\left(\dfrac{1}{3}\right)}{0.05} \\=-\dfrac{ln\left(\dfrac{1}{3}\right)}{0.05}\\t=-20ln\left(\dfrac{1}{3}\right)$

Therefore, the time for which the area will be 15 sqyare kilometers is:

-20 ln(1/3) years.