Question

Solve (a) \(\ln x=2.4050\) (b) \(\ln x=0.9611\) (c) \(\ln x=-0.9611\) (d) \(\ln x=-2.0000\)

Short answer

Question: Find the value of x for each natural logarithm equation. (a) \(\ln x = 2.4050\) (b) \(\ln x = 0.9611\) (c) \(\ln x = -0.9611\) (d) \(\ln x = -2.0000\) Answer: (a) \(x \approx 11.077\) (b) \(x \approx 2.616\) (c) \(x \approx 0.382\) (d) \(x \approx 0.135\)

Step-by-step solution

Understand the given equations

In the exercise, we are given four equations, each containing a natural logarithm of the variable x with a specific value. Our goal is to find the value of x for each equation.

Convert the equation to exponential form

To solve for x, we will convert each natural logarithm equation into an exponential form. The general formula for this conversion is: $$\ln(x) = a \Rightarrow x = e^a$$ We will apply this formula to each equation in the exercise.

Solve (a) \(\ln x=2.4050\)

Using the formula to convert to exponential form and solve for x: $$x = e^{2.4050}$$ Now, compute the value of x: $$x \approx 11.077$$

Solve (b) \(\ln x=0.9611\)

Again, use the formula to convert to exponential form and solve for x: $$x = e^{0.9611}$$ Now, compute the value of x: $$x \approx 2.616$$

Solve (c) \(\ln x=-0.9611\)

Use the formula to convert to exponential form and solve for x: $$x = e^{-0.9611}$$ Now, compute the value of x: $$x \approx 0.382$$

Solve (d) \(\ln x=-2.0000\)

Finally, use the formula to convert to exponential form and solve for x: $$x = e^{-2.0000}$$ Now, compute the value of x: $$x \approx 0.135$$ The solutions for the given equations are as follows: (a) \(x \approx 11.077\) (b) \(x \approx 2.616\) (c) \(x \approx 0.382\) (d) \(x \approx 0.135\)

Key concepts

Exponential Equations

Exponential equations are equations in which variables appear as an exponent. In these problems, the main challenge is to solve for the variable when it is a part of the exponent. One common form is when exponential equations appear in the context of natural logarithms, like \ln(x) = a\. To solve these, converting the logarithmic form into its corresponding exponential form allows you to find the solution by calculating computations more directly.
\[\ln(x) = a \implies x = e^a\]
This conversion is crucial because it leverages the unique properties of logarithms and exponentials, which are inverse operations. By writing the equation in exponential form, you can isolate the variable easily, turning a complex-looking problem into a straightforward computation.

Converting Logarithms

Converting natural logarithms into exponential expressions is an essential skill in algebra. The natural logarithm function \ln(x)\ is the inverse of the exponential function \e^x\. So, if \ln(x) = a\, you can write this in exponential form as \(x = e^a\).
This conversion process is highly useful because it transforms the problem into a form that is often more familiar and computationally manageable. It allows the direct application of exponential rules to compute values, thus solving for x efficiently. In cases where the exponential expression is simple, this conversion step may immediately yield a numerical solution that is easy to approximate using a calculator.

• Always remember that converting logarithms reverses the effect of taking a logarithm, by expressing the problem as a power of e.
• This technique is widely used in calculus and real-world applications, including growth and decay problems.

Solving Equations

When solving equations, especially those involving logarithms and exponentials, the key aim is to isolate the variable of interest—in this instance, x. Starting with a natural logarithm equation, first convert it into exponential form. This conversion simplifies equations as noted in prior sections.
Once in exponential form, compute the value using a calculator to get an approximate solution:

• Understand that every operation on an equation is aimed at simplifying the equation.
• After conversion, solve the equation by calculating the exponential value of expressed powers.
• Verification of your solution can be done by substituting back into the original logarithm equation to see if the left side (the logarithm part) equals the right side (the given value of logarithm).

Solving equations is a fundamental technique in mathematics, helping to unlock solutions to more complex real-world problems like measuring exponential growth in finance or population studies.