Solve for $t$.
$e^{-0.79 t}=0.13$
Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A. $t=$ $\square$
(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 − 0.79 t ) = ln ( 0.13 ) .
Simplify using logarithm properties: − 0.79 t = ln ( 0.13 ) .
Isolate t : t = − 0.79 l n ( 0.13 ) .
Calculate and round: t ≈ 2.583 .
Explanation
Problem Analysis We are given the equation e − 0.79 t = 0.13 and asked to solve for t . The variable t is in the exponent of an exponential function.
Isolating t To solve for t , we need to isolate it. We can do this by taking the natural logarithm of both sides of the equation. The natural logarithm is the logarithm to the base e , and it is denoted as ln . Taking the natural logarithm of both sides gives us: ln ( e − 0.79 t ) = ln ( 0.13 ) Using the property of logarithms that ln ( e x ) = x , we can simplify the left side: − 0.79 t = ln ( 0.13 ) Now, we divide both sides by − 0.79 to isolate t : t = − 0.79 ln ( 0.13 )
Calculating t Now we calculate the value of t :
t = − 0.79 ln ( 0.13 ) ≈ − 0.79 − 2.04022 ≈ 2.582558 Rounding to three decimal places, we get t ≈ 2.583 .
Final Answer Therefore, the solution for t is approximately 2.583 .
Examples
Exponential decay problems, like the one we just solved, are useful in various real-world scenarios. For example, they can model the decay of radioactive substances, the cooling of an object, or the depreciation of an asset. Imagine you bought a new car, and its value depreciates exponentially over time. Solving similar equations would help you determine how long it takes for the car's value to fall below a certain threshold, which is useful for insurance or resale purposes. Understanding exponential decay is also crucial in fields like finance, physics, and engineering.
To solve for t in the equation e − 0.79 t = 0.13 , we take the natural logarithm of both sides, which leads us to the equation t = − 0.79 l n ( 0.13 ) . Calculating this gives t ≈ 2.583 .
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