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Magazine Chapter 4: Problem 47
Find \(f(9)\) if \(f(x)=e^{k x}\) and \(f(3)=2\).
Short Answer
Expert verified
8
Step by step solution
01
- Understand the Function
The given function is of the form: \(f(x) = e^{kx}\)where \(e\) is the base of the natural logarithm, \(k\) is a constant, and \(x\) is the variable.
02
- Substitute Known Values
Use the given value \(f(3) = 2\) to find the constant \(k\). Substitute \(3\) for \(x\) and \(2\) for \(f(x)\): \(2 = e^{3k}\).
03
- Solve for the Constant \(k\)
To solve for \(k\), take the natural logarithm of both sides of the equation \(2 = e^{3k}\): \(\text{ln}(2) = \text{ln}(e^{3k})\).Using the property \(\text{ln}(e^{a}) = a\), we get:\(\text{ln}(2) = 3k\).Finally, solve for \(k\): \(k = \frac{\text{ln}(2)}{3}\).
04
- Rewrite the Function with \(k\)
Now substitute \(k\) back into the original function:\(f(x) = e^{\frac{\text{ln}(2)}{3} x}\).
05
- Find \(f(9)\)
Substitute \(9\) for \(x\) in the function to find \(f(9)\):\(f(9) = e^{\frac{\text{ln}(2)}{3} \times 9}\).Simplify the exponent:\(f(9) = e^{3 \times \text{ln}(2)} = e^{\text{ln}(2^3)} = e^{\text{ln}(8)}\).Since \(e^{\text{ln}(a)} = a\), we get:\(f(9) = 8\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Introduction to Natural Logarithms
In exponential functions, we often encounter the term 'natural logarithm,' often abbreviated as ln. The natural logarithm is the logarithm to the base e, where e is approximately equal to 2.71828.
Unlike common logarithms, which use 10 as the base, the natural logarithm is essential in calculus and various scientific disciplines due to its unique properties.
For example, one critical property we use in solving exponential equations is: \(\text{ln}(e^a) = a\).
This simplification is particularly useful, especially when our function involves exponential terms like the one in the problem we are working with. When solving for a constant embedded in an exponential term, this property allows us to transform a potentially complex equation into a simpler linear form.
The natural logarithm is crucial for simplifying equations and solving for variables hidden within exponential functions.
Unlike common logarithms, which use 10 as the base, the natural logarithm is essential in calculus and various scientific disciplines due to its unique properties.
For example, one critical property we use in solving exponential equations is: \(\text{ln}(e^a) = a\).
This simplification is particularly useful, especially when our function involves exponential terms like the one in the problem we are working with. When solving for a constant embedded in an exponential term, this property allows us to transform a potentially complex equation into a simpler linear form.
The natural logarithm is crucial for simplifying equations and solving for variables hidden within exponential functions.
Solving for the Constant
When solving exponential equations, we often need to find an unknown constant. In our problem, the given function is \(f(x) = e^{kx}\), and we are asked to find \(k\) when \(f(3) = 2\).
To do this, we start by substituting these known values into the equation, obtaining \(2 = e^{3k}\).
To isolate \(k\), we utilize the natural logarithm. By taking the natural logarithm of both sides, we get: \(\text{ln}(2) = \text{ln}(e^{3k})\).
Using the property of natural logarithms, \(\text{ln}(e^a) = a\), the equation simplifies further to: \(\text{ln}(2) = 3k\).
Finally, solve for \(k\) by dividing both sides by 3: \(k = \frac{\text{ln}(2)}{3}\).
This process shows how logarithms can drastically simplify solving for variables in exponential equations, making them more manageable.
To do this, we start by substituting these known values into the equation, obtaining \(2 = e^{3k}\).
To isolate \(k\), we utilize the natural logarithm. By taking the natural logarithm of both sides, we get: \(\text{ln}(2) = \text{ln}(e^{3k})\).
Using the property of natural logarithms, \(\text{ln}(e^a) = a\), the equation simplifies further to: \(\text{ln}(2) = 3k\).
Finally, solve for \(k\) by dividing both sides by 3: \(k = \frac{\text{ln}(2)}{3}\).
This process shows how logarithms can drastically simplify solving for variables in exponential equations, making them more manageable.
Substitution Method
The substitution method involves replacing a variable with a known value to simplify an equation and solve for unknowns.
In our problem, we apply this technique several times. First, we substitute \(x = 3\) and \(f(x) = 2\) into \(f(x) = e^{kx}\) to form the equation \(2 = e^{3k}\).
Next, once we have solved for \(k\), we substitute this value back into the original function to rewrite it in terms of a known constant \(k\).
This gives us: \(f(x) = e^{\frac{\text{ln}(2)}{3} x}\).
Finally, to find \(f(9)\), we substitute 9 for \(x\): \(f(9) = e^{\frac{\text{ln}(2)}{3} \times 9}\).
The substitution method ensures that each step builds upon the previous one, making complex equations easier to solve systematically.
In our problem, we apply this technique several times. First, we substitute \(x = 3\) and \(f(x) = 2\) into \(f(x) = e^{kx}\) to form the equation \(2 = e^{3k}\).
Next, once we have solved for \(k\), we substitute this value back into the original function to rewrite it in terms of a known constant \(k\).
This gives us: \(f(x) = e^{\frac{\text{ln}(2)}{3} x}\).
Finally, to find \(f(9)\), we substitute 9 for \(x\): \(f(9) = e^{\frac{\text{ln}(2)}{3} \times 9}\).
The substitution method ensures that each step builds upon the previous one, making complex equations easier to solve systematically.
