Solve for $t$.
$e^{-0.08t}=0.40$
Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A. $t=$
(Type an integer or a decimal. Do not round until the final answer. Then round to three decimal places as needed. Use a comma to separate answers as needed.)
B. The solution is not a real number.
Answer (2)
Take the natural logarithm of both sides: ln ( e − 0.081 t ) = ln ( 0.40 ) .
Simplify using logarithm properties: − 0.081 t = ln ( 0.40 ) .
Solve for t : t = − 0.081 l n ( 0.40 ) .
Calculate and round: t ≈ 11.312 .
Explanation
Understanding the Problem We are given the equation e − 0.081 t = 0.40 and we want to solve for t .
Isolating the Variable The variable t is in the exponent of the exponential function. To isolate t , we will take the natural logarithm of both sides of the equation.
Applying Logarithms and Solving for t Taking the natural logarithm of both sides, we have: ln ( e − 0.081 t ) = ln ( 0.40 ) Using the property of logarithms that ln ( e x ) = x , we simplify the left side: − 0.081 t = ln ( 0.40 ) Now, we divide both sides by − 0.081 to solve for t :
t = − 0.081 ln ( 0.40 )
Calculating the Value of t Calculating the value of t , we get: t = − 0.081 ln ( 0.40 ) ≈ 11.312 Rounding to three decimal places, we have t ≈ 11.312 .
Final Answer Therefore, the solution is t ≈ 11.312 .
Examples
Exponential decay is a common phenomenon in various fields. For instance, in finance, the value of an asset may decrease exponentially over time due to depreciation. Similarly, in physics, radioactive decay follows an exponential pattern. Understanding how to solve equations involving exponential decay allows us to model and predict these real-world phenomena accurately. For example, if a car's value depreciates at a rate of 8.1% per year and is currently worth 25 , 000 , w ec an u se t h ee q u a t i o n V(t) = 25000e^{-0.081t}$ to determine when the car's value will be $10,000.
The solution to the equation e − 0.08 t = 0.40 is t ≈ 11.454 after applying the natural logarithm and isolating t .
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