Problem 76 - College Algebra and Calculus: An Applied Approach

Question

Solve the logarithmic equation algebraically. Approximate the result to three decimal places.\(2-6 \ln x=10\)

Step-by-Step Solution

Answer

The approximate solution to the equation \(2-6 \ln x=10\) is \(x \approx 0.263\).

Step 1: Isolate the Natural Logarithm

Start by isolating the log term by subtracting \(2\) from each side of the equation, we get: \(-6 \ln x = 10 - 2\), which simplifies to \(-6 \ln x = 8\).

Step 2: Remove the coefficient from the natural logarithm

Divide both sides by \(-6\) to remove the coefficient from the natural logarithm, thus getting: \(\ln x = -\frac{4}{3}\).

Step 3: Remove the logarithm

To remove the log from the equation and isolate \(x\), use the property that says if \(\ln a = b\), then \(a = e^b\). Here, \(e^{-\frac{4}{3}}=x\).

Step 4: Approximate

Now, compute \(e^{-\frac{4}{3}}\) using a calculator and round to three decimal places. After this step, we get \(x \approx 0.263\).

Key Concepts

Natural Logarithms

Natural logarithms are logarithms that use the base of the mathematical constant \( e \), which is approximately equal to 2.71828. These logarithms are denoted as \( \ln \). When we encounter an equation involving a natural logarithm, such as \( \ln x = y \), we can rewrite it using its exponential form: \( x = e^y \). This transformation allows us to easily convert logarithmic equations into exponential ones, making it simpler to solve for the variable.

• Base \( e \): The natural number \( e \) is fundamental in mathematics and arises naturally in processes that show continuous growth, like compound interest or population growth.


• Using \( \ln \): When you see \( \ln \) in an equation, it indicates that the base of the logarithm is \( e \). If you isolate the \( \ln \) term, you can convert it to an exponential form.


• Application: Knowing how to manipulate \( \ln \) helps in solving equations that model exponential growth or decay, common in physics and finance.

Exponential Functions

Exponential functions involve expressions where variables appear in the exponent, rather than the base. In the equation \( \ln x = -\frac{4}{3} \), we rearrange this to find \( x = e^{-\frac{4}{3}} \).

• Understanding \( e^x \): Exponential expressions with base \( e \) can model continuous growth or decay. This property is why \( e \) is so useful and ubiquitous in mathematics.


• Conversions: You might need to solve equations by converting from a logarithmic form to an exponential one. This method often reveals the solution more clearly and concisely.


• Computation: Calculating the value of an exponential expression, such as \( e^{-\frac{4}{3}} \), helps find exact values that can be approximated to the desired precision.

Algebraic Manipulation

Algebraic manipulation involves rearranging parts of an equation to isolate the variable of interest. In the original equation \( 2 - 6 \ln x = 10 \), several algebraic steps were taken to solve for \( x \).

• Isolation: Subtracting and dividing are key algebraic tools used to isolate \( \ln x \). Every operation applied to one side of an equation must be similarly applied to the other to keep the equation balanced.


• Coefficient Removal: Dealing with coefficients attached to \( \ln x \), such as \(-6\), involves dividing both sides of the equation, effectively "smoothing" the path to solve for the variable.


• Precision: To find a decimal approximation, use appropriate resources like a calculator. Rounding helps convey solutions clearly, seen here when approximating \( e^{-\frac{4}{3}} \) to 0.263.