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Answer :
Sure! Let's transform the given exponential function into the specified form step by step.
We are given the function:
[tex]\[ f(t) = 60000 \cdot e^{-0.1t} \][/tex]
And we need to write it in the form:
[tex]\[ f(t) = a\left(\frac{1}{2}\right)^{\frac{t}{k}} \][/tex]
Step 1: Identify the coefficient [tex]\( a \)[/tex].
In both forms of the function, [tex]\( a \)[/tex] represents the initial value when [tex]\( t = 0 \)[/tex]. For the given function, you can see that when [tex]\( t = 0 \)[/tex]:
[tex]\[ f(0) = 60000 \cdot e^0 = 60000 \][/tex]
Thus, [tex]\( a = 60000 \)[/tex].
Step 2: Find the value of [tex]\( k \)[/tex].
We want the expressions to be equal, which means finding [tex]\( k \)[/tex] such that:
[tex]\[ e^{-0.1t} = \left(\frac{1}{2}\right)^{\frac{t}{k}} \][/tex]
Taking the natural logarithm (base [tex]\( e \)[/tex]) of both sides will help us to solve for [tex]\( k \)[/tex]:
[tex]\[ -0.1t = \frac{t}{k} \ln\left(\frac{1}{2}\right) \][/tex]
We know that [tex]\( \ln\left(\frac{1}{2}\right) \)[/tex] is a negative value, so this equation will effectively help us equate the two expressions. Solving for [tex]\( k \)[/tex]:
[tex]\[ -0.1 = \frac{\ln(0.5)}{k} \][/tex]
Now, rearrange to solve for [tex]\( k \)[/tex]:
[tex]\[ k = \frac{\ln(0.5)}{-0.1} \][/tex]
By computing this, we find that:
[tex]\[ k \approx 14.4269 \][/tex]
Step 3: Write the final function.
Now we can substitute [tex]\( a \)[/tex] and [tex]\( k \)[/tex] back into the expression to get:
[tex]\[ f(t) = 60000 \left(\frac{1}{2}\right)^{\frac{t}{14.4269}} \][/tex]
And that's the equivalent form of the given function!
We are given the function:
[tex]\[ f(t) = 60000 \cdot e^{-0.1t} \][/tex]
And we need to write it in the form:
[tex]\[ f(t) = a\left(\frac{1}{2}\right)^{\frac{t}{k}} \][/tex]
Step 1: Identify the coefficient [tex]\( a \)[/tex].
In both forms of the function, [tex]\( a \)[/tex] represents the initial value when [tex]\( t = 0 \)[/tex]. For the given function, you can see that when [tex]\( t = 0 \)[/tex]:
[tex]\[ f(0) = 60000 \cdot e^0 = 60000 \][/tex]
Thus, [tex]\( a = 60000 \)[/tex].
Step 2: Find the value of [tex]\( k \)[/tex].
We want the expressions to be equal, which means finding [tex]\( k \)[/tex] such that:
[tex]\[ e^{-0.1t} = \left(\frac{1}{2}\right)^{\frac{t}{k}} \][/tex]
Taking the natural logarithm (base [tex]\( e \)[/tex]) of both sides will help us to solve for [tex]\( k \)[/tex]:
[tex]\[ -0.1t = \frac{t}{k} \ln\left(\frac{1}{2}\right) \][/tex]
We know that [tex]\( \ln\left(\frac{1}{2}\right) \)[/tex] is a negative value, so this equation will effectively help us equate the two expressions. Solving for [tex]\( k \)[/tex]:
[tex]\[ -0.1 = \frac{\ln(0.5)}{k} \][/tex]
Now, rearrange to solve for [tex]\( k \)[/tex]:
[tex]\[ k = \frac{\ln(0.5)}{-0.1} \][/tex]
By computing this, we find that:
[tex]\[ k \approx 14.4269 \][/tex]
Step 3: Write the final function.
Now we can substitute [tex]\( a \)[/tex] and [tex]\( k \)[/tex] back into the expression to get:
[tex]\[ f(t) = 60000 \left(\frac{1}{2}\right)^{\frac{t}{14.4269}} \][/tex]
And that's the equivalent form of the given function!
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Rewritten by : Barada
