A large open meadow near your city is shrinking in size because developers have started building new homes there.
The relationship between [tex]$A$[/tex], the area of the meadow, in hectares, and [tex]$t$[/tex], the elapsed time, in months, since the construction began is modeled by the following function.
[tex]$A=1562.5 \cdot 10^{-0.1 t}$[/tex]
How many months of construction will there be before the area of the meadow decreases to 500 hectares?
Give an exact answer expressed as a base-10 logarithm.
[ ] months
Answer (1)
Substitute A = 500 into the equation A = 1562.5 \tcdot 1 0 − 0.1 t .
Simplify the equation to isolate the exponential term: 0.32 = 1 0 − 0.1 t .
Take the base-10 logarithm of both sides: lo g 10 ( 0.32 ) = − 0.1 t .
Solve for t : t = − 10 lo g 10 ( 0.32 ) .
The number of months is − 10 lo g 10 ( 0.32 ) .
Explanation
Understanding the Problem We are given the equation A = 1562.5 \tcdot 1 0 − 0.1 t , where A is the area of the meadow in hectares and t is the time in months since construction began. We want to find the time t when the area A is 500 hectares.
Substituting the Value of A Substitute A = 500 into the equation: 500 = 1562.5 \tcdot 1 0 − 0.1 t
Isolating the Exponential Term Divide both sides by 1562.5: 1562.5 500 = 1 0 − 0.1 t
Simplifying the Fraction Simplify the fraction: 1562.5 500 = 0.32 So we have: 0.32 = 1 0 − 0.1 t
Taking the Base-10 Logarithm Take the base-10 logarithm of both sides: lo g 10 ( 0.32 ) = lo g 10 ( 1 0 − 0.1 t )
Applying Logarithm Properties Use the property of logarithms lo g 10 ( 1 0 x ) = x : lo g 10 ( 0.32 ) = − 0.1 t
Solving for t Divide both sides by -0.1: t = − 0.1 lo g 10 ( 0.32 )
Simplifying the Expression Multiply the numerator and denominator by -1: t = − 10 lo g 10 ( 0.32 )
Final Answer Therefore, the number of months of construction before the area decreases to 500 hectares is: t = − 10 lo g 10 ( 0.32 ) months.
Examples
This type of exponential decay problem can be applied to various real-world scenarios, such as modeling the depreciation of a car's value over time. If a car's initial value is 15625 , an d i t d e p rec ia t es a t a r a t es u c h t ha t i t s v a l u e i s m o d e l e d b y V = 15625 \tcdot 10^{-0.1t} , w h ere V i s t h ec a r ′ s v a l u e an d t$ is the time in years, we can determine how long it will take for the car's value to drop to $500 using the same logarithmic approach. Understanding exponential decay helps in making informed decisions about investments and asset management.
