Problem 65 - College Algebra and Calculus: An Applied Approach

Question

Solve the logarithmic equation algebraically. Approximate the result to three decimal places.\(\ln 2 x=2.4\)

Step-by-Step Solution

Answer

The solution to the equation \(\ln 2x = 2.4\) is approximately \(x = 2.714\) when rounded to three decimal places.

Step 1: Transcribe to Exponential Form

To solve for \(x\), we first need to convert our logarithmic equation into an exponential one. This can be done by using the fact that \(\ln a = b\) is equivalent to \(e^b = a\). Applying this to the given equation \(\ln 2x = 2.4\) transforms it into \(e^{2.4} = 2x\).

Step 2: Solve for \(x\)

We now have the equation \(e^{2.4} = 2x\). We can solve for \(x\) by dividing both sides of the equation by 2 to isolate \(x\). This gives us \(x = e^{2.4}/2\).

Step 3: Compute the Value

Now that we've isolated \(x\), we can use a calculator to work out the value of \(e^{2.4}/2\). Calculating this gives \(x = 5.48/2\), which simplifies to approximately \(x = 2.714\) (rounded to three decimal places).

Key Concepts

From Logarithmic to Exponential Form

Understanding how to switch between logarithmic and exponential forms is crucial when solving equations involving logarithms. The natural logarithm, denoted as \(\ln x\), is a logarithm to the base \(e\), where \(e\) is the mathematical constant approximately equal to 2.718. When you have an equation like \(\ln a = b\), this can be rewritten in its exponential form as \(e^b = a\). This relationship exists because the natural logarithm is the power to which \(e\) must be raised to get the number \(a\).

For example, in the exercise provided, \(\ln 2x = 2.4\) is equivalent to \(e^{2.4} = 2x\) after conversion to exponential form. This step is key in the process of solving logarithmic equations because it transforms the problem into a more familiar algebraic equation.

Natural Logarithm Properties

The natural logarithm has unique properties that make it a valuable tool for solving equations. One such property is that the natural logarithm of \(e\), \(\ln e\), is always 1. This is because \(e\) raised to the power of 1 is \(e\). Another property is that the natural logarithm of 1, \(\ln 1\), is 0 because \(e\) raised to the power of 0 is 1. Moreover, the natural logarithm follows the rules of logarithms, such as:

• Product Rule: \(\ln(ab) = \ln(a) + \ln(b)\)
• Quotient Rule: \(\ln(\frac{a}{b}) = \ln(a) - \ln(b)\)
• Power Rule: \(\ln(a^b) = b\ln(a)\)

These properties are important for manipulating and simplifying logarithmic expressions when isolating variables and solving equations.

Isolating Variables Algebraically

Solving for an unknown variable often involves isolating it on one side of the equation. When dealing with exponential equations, such as those arising from logarithmic equations, isolating the variable requires specific algebraic operations. In the context of the exercise we are examining, once the logarithmic equation is converted into the exponential form (\(e^{2.4} = 2x\)), the next step is to perform algebraic operations that will leave \(x\) by itself on one side of the equation.

Dividing both sides by 2 (\(\frac{e^{2.4}}{2} = x\)) successfully isolates \(x\) and enables you to solve for it. The core concept here is to use inverse operations to undo the operations being performed on the variable. In the case of multiplication by 2, division by 2 is the inverse operation necessary to isolate \(x\). By mastering algebraic techniques to isolate variables, one can solve a wide range of mathematical problems.