Problem 47 - A Graphical Approach to College Algebra

Question

Solve each logarithmic equation. Express all solutions in exact form. Support your solutions by using a calculator. $$5 \log \left(x^{2}-1\right)+7=12$$

Step-by-Step Solution

Answer

The solutions are \( x = \pm \sqrt{11} \).

Step 1: Isolate the Logarithmic Term

We start by isolating the logarithmic term on one side of the equation. Subtract 7 from both sides of the equation:\[ 5 \log(x^2 - 1) + 7 - 7 = 12 - 7 \]Simplifying both sides gives:\[ 5 \log(x^2 - 1) = 5 \]

Step 2: Divide by the Coefficient of the Logarithm

To solve for the logarithmic term, divide both sides by 5 to remove the coefficient:\[ \frac{5 \log(x^2 - 1)}{5} = \frac{5}{5} \]This simplifies to:\[ \log(x^2 - 1) = 1 \]

Step 3: Convert to Exponential Form

Convert the logarithmic equation to its equivalent exponential form. Recall that \( \log_b(a) = c \) means \( b^c = a \). Here, the base is 10 by default:\[ 10^1 = x^2 - 1 \]This simplifies to:\[ 10 = x^2 - 1 \]

Step 4: Solve the Quadratic Equation

Rearrange the equation to solve for \( x^2 \):\[ x^2 = 10 + 1 \]Thus, \( x^2 = 11 \).To solve for \( x \), take the square root of both sides:\[ x = \pm \sqrt{11} \]

Step 5: Verify the Solutions

Verify if both solutions satisfy the original logarithmic condition (i.e., \( x^2 - 1 > 0 \), since the argument of a log must be positive). Check:For \( x = \sqrt{11} \):- \( x^2 = 11 \) implies \( 11 - 1 = 10 > 0 \).For \( x = -\sqrt{11} \):- \( x^2 = 11 \) also implies \( 11 - 1 = 10 > 0 \).Both satisfy the conditions, so both values are valid solutions.

Key Concepts

Exponential Form Conversion

Logarithmic equations often require conversion from the logarithmic form to the exponential form to make them easier to solve. Remember:

• In a logarithmic equation like \( \log_b(a) = c \), the exponential form is \( b^c = a \).
• The base \( b \) is crucial here, and common bases include 10 and \( e \) (the natural logarithm base).

In the given problem, we had \( \log(x^2 - 1) = 1 \). This equation uses the common logarithm, which has a base of 10 by default. Thus, we convert it:

• Exponentially: \( 10^1 = x^2 - 1 \), leading to \( x^2 - 1 = 10 \).
• Solving this gives us our path to find \( x \).

This conversion simplifies solving logarithmic equations by bringing them into a more familiar form, similar to a basic algebraic equation.

Quadratic Equations

Quadratic equations are polynomial equations of the second degree, generally in the form \( ax^2 + bx + c = 0 \). Solving them typically involves:

• Rearranging the equation into standard form.
• Using perfect squares, factoring, or the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

In our exercise, once we rearranged \( x^2 - 1 = 10 \) to \( x^2 = 11 \), it naturally set us up for solving a quadratic equation. Since there are no linear \( x \) terms or other factors, we simply took the square root of both sides:

• \( x = \pm \sqrt{11} \).

It's crucial to consider both the positive and negative roots in such cases, as both are valid solutions unless further restrictions (such as domain) limit them.

Logarithm Properties

Logarithms have distinct properties that are highly useful in simplifying equations and solving problems:

• The equation \( \log_b(a) = c \) implies \( b^c = a \), as seen in exponential conversion.
• They can only accept positive arguments, which is why we check conditions like \( x^2 - 1 > 0 \).
• Key properties include using \( \, \log_b(mn) = \log_b(m) + \log_b(n) \, \) and \( \, \log_b(\frac{m}{n}) = \log_b(m) - \log_b(n) \, \).

In our exercise, the property that the argument \((x^2 - 1) \) must remain positive was crucial after we solved for \( x \). Ensuring this condition helped us appreciate that even negative square roots may lead to valid solutions when squared, provided they meet the property conditions of logarithms.Understanding these properties aids significantly in working through problems systematically and recognizing viable solutions.