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

To find the number of years it will take for the glacier's area to decrease to 15 square kilometers, the equation A=45e^{-0.05t} is solved for t, giving t = \frac{ln(\frac{1}{3})}{-0.05}.

Step-by-step explanation:

The student is asked to calculate the time t it will take for the area of a glacier, initially covering 454545 square kilometers, to decrease to 151515 square kilometers. The area A as a function of time t is given by the exponential decay function A=45e^{-0.05t}. To find t when A is 15, we set A=15 and solve for t using the natural logarithm:

• Set the equation: 15=45e^{-0.05t}
• Divide both sides by 45: \frac{15}{45}=e^{-0.05t}
• Simplify: \frac{1}{3}=e^{-0.05t}
• Take the natural logarithm of both sides: ln(\frac{1}{3}) = -0.05t
• Divide by -0.05: t = \frac{ln(\frac{1}{3})}{-0.05}

The exact time t for the glacier's area to decrease to 15 square kilometers is expressed as t = \frac{ln(\frac{1}{3})}{-0.05}.

Answer 2

Answer: t=-20ln(1/3)

Step-by-step explanation: