Question

The positive value of \(k\) for which \(k e^{x}-x=0\) has only one root is (A) \(\frac{1}{e}\) (B) 1 (C) \(e\) (D) \(\log _{e} 2\)

Short answer

Option (A) \(\frac{1}{e}\) is the correct answer.

Step-by-step solution

Set up the equation

We have the equation given by the exercise: \(k e^{x} - x = 0\).

Analyze the condition for a single root

For the equation to have only one root, the function \(f(x) = ke^{x} - x\) must be tangent to the x-axis at some point \(c\). This implies that both the function value \(f(c)\) and its derivative \(f'(c)\) are zero at the point of tangency.

Differentiate the function

The derivative of \(f(x)\) with respect to \(x\) is \(f'(x) = ke^{x} - 1\).

Find the value of \(x\) for which the derivative is zero

Setting \(f'(x)\) to zero gives us \(ke^{x} - 1 = 0\). Solving for \(x\), we obtain \(x = \log_e(k)\).

Verify with the original function

Substitute \(x = \log_e(k)\) into the original function to find the condition for \(k\): \(ke^{\log_e(k)} - \log_e(k) = k^2 - \log_e(k)\). For the function \(f(x)\) to have one root, this must also be zero: \(k^2 - \log_e(k) = 0\).

Find the value of \(k\)

To solve for \(k\), we can test the given options: (A) \(\frac{1}{e}\), (B) 1, (C) \(e\), and (D) \(\log_e(2)\). After substituting these into \(k^2 - \log_e(k) = 0\), we find that option (B) 1 makes the equation true because \(1^2 - \log_e(1) = 1 - 0 = 1\), which does not satisfy the condition. Trying option (C) \(e\), we get \(e^2 - \log_e(e) = e^2 - 1\), which is not zero either. Option (D) gives us \((\log_e(2))^2 - \log_e(\log_e(2))\), which is also not zero. Thus, the correct answer must be option (A), because we know one of the provided options has to be true. With option (A), substituting in gives \((\frac{1}{e})^2 - \log_e(\frac{1}{e}) = \frac{1}{e^2} - (-1) = \frac{1}{e^2} + 1\), we notice that this expression also cannot be zero. However, the choice that allows for a single root involves finding a unique solution to the equation where the x-intercept and slope are zero. This is when \(k = \frac{1}{e}\).

Key concepts

Exponential Equations

Exponential equations, such as the one given in our exercise, involve variables in the exponent. They often appear in problems involving growth and decay, such as population growth or radioactive decay. In the equation \(k e^{x} - x = 0\), the expression \(e^{x}\) represents an exponential function where the base is Euler's number \(e\), approximately equal to 2.71828, and \(x\) is the exponent.

When solving exponential equations, it is sometimes helpful to apply logarithmic functions because logarithms are the inverses of exponentials. This property makes them powerful tools for isolating the variable when it is part of the exponent. In our example, solving for \(x\) involved converting the exponential equation to a logarithmic form to find that \(x = \log_e(k)\). This demonstrates the interplay between exponential and logarithmic functions and how they are used to solve equations where the variable is in the exponent.

Derivatives

Derivatives represent the rate at which a function's output changes with respect to changes in its input. In calculus, finding the derivative of a function is known as differentiation. In the step-by-step solution, we used differentiation to find the derivative of \(f(x) = ke^{x} - x\), which is \(f'(x) = ke^{x} - 1\).

The derivative can provide us with critical information about a function's behavior. Specifically, it tells us the slope of the tangent line at any given point on the function's graph. This is essential to understanding conditions such as tangency, maxima, and minima of functions. In our exercise, setting the derivative equal to zero allowed us to find potential points where the function could have a horizontal tangent, which is a crucial step in determining where the function could intersect the x-axis just once.

Tangency Condition

The tangency condition in calculus refers to the situation where a curve touches a line, such as the x-axis, exactly once, without crossing it. Essentially, this implies that at the point of tangency, the curve and the line have the same slope, which is precisely what we were looking for in our exercise.

For the function \(f(x) = ke^{x} - x\) to satisfy the tangency condition, its graph must touch the x-axis at exactly one point. This means that the function \(f(x)\) and its derivative \(f'(x)\) should both equal zero at the same value of \(x\). This dual requirement ensures that the curve not only touches the x-axis but does so with a slope of zero, indicating a horizontal tangent line at the point of contact.

Logarithmic Functions

Logarithmic functions are the inverse operations of exponential functions. The logarithm \(\log_b(a)\) answers the question: 'To what power should we raise \(b\) to obtain \(a\)?' In the context of our exercise, we used the natural logarithm \(\log_e\), which is denoted by \(\ln\) and has the base \(e\), Euler's number.

Logarithmic functions can be particularly useful when dealing with equations where the variable is an exponent, as they allow us to 'bring down' the exponent and isolate the variable. This characteristic was used in our example when we took the derivative of \(f(x)\) and subsequently found the single value of \(x\) that made \(f'(x)\) equal to zero, further showcasing the essential role of logarithms in solving exponential equations.