Solve for $t$.
$e^{-0.24 t}=0.59$
Answer (2)
Take the natural logarithm of both sides: ln ( e − 0.24 t ) = ln ( 0.59 ) .
Simplify using logarithm properties: − 0.24 t = ln ( 0.59 ) .
Calculate the natural logarithm: − 0.24 t = − 0.5276 .
Solve for t : t = − 0.24 − 0.5276 ≈ 2.198 .
The solution is 2.198 .
Explanation
Problem Analysis We are given the equation e − 0.24 t = 0.59 and we want to solve for t . The variable t is in the exponent of the exponential function.
Solution Strategy To solve for t , we need to isolate it. Since t is in the exponent, we can take the natural logarithm of both sides of the equation. This will allow us to use the property of logarithms that ln ( e x ) = x .
Applying Natural Logarithm Taking the natural logarithm of both sides of the equation e − 0.24 t = 0.59 , we get: ln ( e − 0.24 t ) = ln ( 0.59 ) Using the property of logarithms, we have: − 0.24 t = ln ( 0.59 ) From calculation, we have ln ( 0.59 ) ≈ − 0.527632742082372 . Thus, − 0.24 t = − 0.527632742082372
Isolating t Now, we isolate t by dividing both sides of the equation by − 0.24 :
t = − 0.24 − 0.527632742082372 t ≈ 2.19846975867655
Final Answer Therefore, the value of t is approximately 2.198 .
Examples
Exponential decay is a mathematical concept with many real-world applications. For example, it can model the cooling of an object. Imagine you have a cup of hot coffee, and you want to know how long it will take to cool down to a drinkable temperature. The temperature of the coffee decreases exponentially with time, and the equation T ( t ) = T s + ( T 0 − T s ) e − k t models this, where T ( t ) is the temperature at time t , T s is the surrounding temperature, T 0 is the initial temperature, and k is a constant. Solving for t in such equations helps determine how long to wait before the coffee is cool enough to drink.
To solve the equation e − 0.24 t = 0.59 , we take the natural logarithm of both sides, simplify, and then isolate t to find that t ≈ 2.198 . The steps involve using properties of logarithms and basic algebraic manipulation.
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