Respuesta :
We have been given that a huge ice glacier in the Himalayas initially covered an area of 45 square kilo-meters. The relationship between A, the area of the glacier in square kilo-meters, and t, the number of years the glacier has been melting, is modeled by the equation [tex]A=45e^{-0.05t}[/tex].
To find the time it will take for the area of the glacier to decrease to 15 square kilo-meters, we will equate [tex]A=15[/tex] and solve for t as:
[tex]15=45e^{-0.05t}[/tex]
[tex]\frac{15}{45}=\frac{45e^{-0.05t}}{45}[/tex]
[tex]\frac{1}{3}=e^{-0.05t}[/tex]
Now we will switch sides:
[tex]e^{-0.05t}=\frac{1}{3}[/tex]
Let us take natural log on both sides of equation.
[tex]\text{ln}(e^{-0.05t})=\text{ln}(\frac{1}{3})[/tex]
Using natural log property [tex]\text{ln}(a^b)=b\cdot \text{ln}(a)[/tex], we will get:
[tex]-0.05t\cdot \text{ln}(e)=\text{ln}(\frac{1}{3})[/tex]
[tex]-0.05t\cdot (1)=\text{ln}(\frac{1}{3})[/tex]
[tex]-0.05t=\text{ln}(\frac{1}{3})[/tex]
[tex]t=\frac{\text{ln}(\frac{1}{3})}{-0.05}[/tex]
[tex]t=\frac{\text{ln}(\frac{1}{3})\cdot 100}{-0.05\cdot 100}[/tex]
[tex]t=\frac{\text{ln}(\frac{1}{3})\cdot 100}{-5}[/tex]
[tex]t=-\text{ln}(\frac{1}{3})\cdot 20[/tex]
[tex]t=-(\text{ln}(1)-\text{ln}(3))\cdot 20[/tex]
[tex]t=-(0-\text{ln}(3))\cdot 20[/tex]
[tex]t=20\text{ln}(3)[/tex]
Therefore, it will take [tex]20\text{ln}(3)[/tex] years for area of the glacier to decrease to 15 square kilo-meters.
