Answer:
B. 7.3 years
Step-by-step explanation:
The question is wrong. The correct function is :
[tex]g(x)=12500(0.91)^{x}[/tex]
We have the function [tex]g(x)[/tex] that represents the value of a piece of farm equipment after [tex]x[/tex] years.
This means that when [tex]x=0[/tex] its original value is :
[tex]g(0)=12500(0.91)^{0}=12500[/tex]
Now we want to calculate approximately when will its value be half its original value. Then, we write :
[tex]\frac{12500}{2}=6250[/tex]
[tex]6250[/tex] is half of its original value. We need to find [tex]x[/tex] that satisfies the following equation :
[tex]g(x)=6250=12500(0.91)^{x}[/tex]
Solving for [tex]x[/tex] :
[tex]6250=12500(0.91)^{x}[/tex] ⇒
[tex]0.5=(0.91)^{x}[/tex]
Now we apply natural logarithm to each side of the equation :
[tex]ln(0.5)=ln[(0.91)^{x}][/tex]
Using logarithm properties :
[tex]ln(0.5)=ln[(0.91)^{x}][/tex] ⇒
[tex]ln(0.5)=x[ln(0.91)][/tex]
[tex]x=\frac{ln(0.5)}{ln(0.91)}[/tex] ⇒
[tex]x[/tex] ≅ [tex]7.3496[/tex]
The correct option is B. 7.3 years
