Problem 79 - Elementary and Intermediate Algebra

Question

Solve each equation. Give the exact solution and, when appropriate, an approximation to four decimal places. $$ \log _{3} x=\log _{3}\left(\frac{1}{x}\right)+4 $$

Step-by-Step Solution

Answer

The exact solution is \( x = 9 \). Approximation is not needed as the solution is exact.

Step 1: Use Logarithm Properties

The equation given is \( \log_{3}x = \log_{3}\left(\frac{1}{x}\right) + 4 \). By using the property of logarithms that states \( \log_{b}M - \log_{b}N = \log_{b}\left(\frac{M}{N}\right) \), we can rewrite the equation as: \[\log_{3}\left(\frac{x}{\frac{1}{x}}\right) = 4.\] This simplifies to: \(\log_{3}(x^2) = 4\).

Step 2: Remove the Logarithm

To eliminate the logarithm, rewrite the equation in its exponential form: \( x^2 = 3^4 \). We know that \( 3^4 = 81 \). Hence, \( x^2 = 81 \).

Step 3: Solve for x

Taking the square root on both sides, we get \( x = \pm \sqrt{81} \). Since \( \sqrt{81} = 9 \), the solutions are \( x = 9 \) and \( x = -9 \).

Step 4: Check the Solutions

Check whether both solutions are valid in the original logarithmic equation. \( \log_{3}(-9) \) is undefined because logarithms of negative numbers are not real. Therefore, \( x = 9 \) is the only valid solution.

Key Concepts

Logarithm Properties

Logarithms have specific properties that help us simplify and solve equations. One vital property is the difference of logarithms rule:

• \( \log_{b}M - \log_{b}N = \log_{b}\left(\frac{M}{N}\right) \).

This property is particularly useful when you have a subtraction of two logarithms with the same base. It allows you to combine them into a single logarithm, which simplifies the equation.
In our original equation, \( \log_{3}x = \log_{3}\left(\frac{1}{x}\right) + 4 \), realizing this property means we change the equation to \( \log_{3}\left(\frac{x}{\frac{1}{x}}\right) = 4 \).
As a result, the consolidation reduces complexity and lets us work with simpler forms moving forward.

Logarithmic and Exponential Forms

Logarithms and exponents are closely related. A logarithm answers the question: “To what exponent must the base be raised to produce a given number?”. This is crucial for switching between logarithmic and exponential forms.
The key conversion formula is:

• If \( \log_{b}A = C \), then \( b^C = A \).

In our exercise, once simplified to \( \log_{3}(x^2) = 4 \), we can eliminate the logarithm by translating it to its exponential form: \( x^2 = 3^4 \).
This conversion provides a clear path to solving the equation, putting us on solid ground to find the value of \( x \).

Solving Equations

Solving equations involves finding specific values for variables that make the equation true. Once we have \( x^2 = 81 \), the next step is to solve for \( x \) by taking the square root of both sides.
When you solve \( x^2 = 81 \), it results in two potential solutions:

• \( x = 9 \)
• \( x = -9 \)

Both values satisfy this equation because squaring either will result in 81. However, not all solutions will be valid when you substitute them back into the original equation, due to restrictions on logarithmic expressions.

Checking Solutions

Checking solutions ensures that the results fulfill the conditions of the original equation. After finding potential solutions \( x = 9 \) and \( x = -9 \), we must evaluate whether they are valid in the context of the initial logarithmic equation.
Logarithms are only defined for positive numbers. Because \( \log_{3}(-9) \) is undefined, this means \( x = -9 \) is not a valid solution.
Therefore, substituting back into the original equation verifies that \( x = 9 \) is the sole valid solution:

• \( \log_{3}(9) = \log_{3}\left(\frac{1}{9}\right) + 4 \) reduces to a statement that holds true.

Thus, checking each solution by substituting it back into the original equation confirms whether it is correct.