Problem 15 - Precalculus

Question

Solve the equation analytically. $$ \log _{3}(x-4)+\log _{3}(x+4)=2 $$

Step-by-Step Solution

Answer

The solution is \(x = 5\).

Step 1: Apply the Logarithm Property

The equation starts with the sum of two logarithms, which can be rewritten as a single logarithm using the property \(\log _{b}(m) + \log _{b}(n) = \log _{b}(mn)\). So, \(\log _{3}(x-4) + \log _{3}(x+4) = \log _{3}((x-4)(x+4))\).

Step 2: Simplify the Expression

The expression \((x-4)(x+4)\) can be expanded using the difference of squares formula, \((a-b)(a+b) = a^2-b^2\). Thus, \((x-4)(x+4) = x^2 - 16\).

Step 3: Set Up the Equation with Logarithms

Now, substitute \(x^2 - 16\) into the equation from Step 1: \(\log _{3}(x^2 - 16) = 2\).

Step 4: Convert Logarithmic to Exponential Form

Convert the logarithmic equation \(\log _{3}(x^2 - 16) = 2\) into exponential form: \(x^2 - 16 = 3^2\).

Step 5: Solve the Exponential Equation

Calculate \(3^2\), which is 9. Then the equation becomes \(x^2 - 16 = 9\). Add 16 to both sides to obtain \(x^2 = 25\).

Step 6: Solve for x

Take the square root of both sides to solve for \(x\): \(x = \pm \sqrt{25}\), which gives \(x = 5\) or \(x = -5\).

Step 7: Check for Valid Solutions

Substitute \(x = 5\) back into the original logarithmic expressions: \(\log _{3}(5-4) + \log _{3}(5+4) = \log _{3}(1) + \log _{3}(9)\), which equals \(0 + 2 = 2\), matching the right side. For \(x = -5\), \(\log _{3}(-5-4)\) is undefined, so \(x = -5\) is not a valid solution.

Key Concepts

Logarithm Properties

Logarithms are a powerful way to work with exponential equations by transforming multiplicative processes into additive ones. A key property of logarithms is that the logarithm of a product can be expressed as the sum of the logarithms of the factors. This is formulated as:

• \( \log_b(mn) = \log_b(m) + \log_b(n) \)

This property allows us to combine or simplify logarithmic expressions.
In our exercise, we used this property to combine two logarithmic expressions: \( \log_3(x-4) + \log_3(x+4) \). By using the property, we rewrote this as:

• \( \log_3((x-4)(x+4)) \)

This simplification makes it easier to work with the equation analytically.

Difference of Squares

The difference of squares is a special algebraic pattern that involves expressing products of two binomials as

• \( (a-b)(a+b) = a^2 - b^2 \)

This formula plays a significant role in algebraic simplification and helps make complex expressions more manageable.
In our problem, the expression \((x-4)(x+4)\) is expanded using this property. By applying the formula, we get:

• \( x^2 - 16 \)

This form is much simpler and directly helped us solve the logarithmic equation by substituting it back easily into the equation.

Exponential Equations

Exponential equations often involve unknown variables in the exponent and can usually be solved by manipulating the expressions to isolate the variable.
In our exercise, after applying logarithmic and algebraic properties, we arrive at a logarithmic equation:

• \( \log_3(x^2 - 16) = 2 \)

This can be transformed into an exponential equation by using the definition of a logarithm. This tells us that:

• \( a = b^c \) is the same as \( \log_b(a) = c \)

Thus, we translate the logarithmic equation into:

• \( x^2 - 16 = 3^2 \)

This sets up the exponential equation to be solved further.

Valid Solutions in Logarithms

When dealing with solutions to logarithmic equations, it's crucial to recognize the domain constraints of logarithm functions. Logarithms are only defined for positive real numbers. This means the argument of a logarithm must always be greater than zero.
During our solution check, we needed to ensure the solutions fit these constraints. We originally had solutions for \(x = 5\) and \(x = -5\). When substituting \(x = 5\) into the logarithms, they are valid since all values become positive:

• \( \log_3(5-4) + \log_3(5+4) = 0 + 2 = 2 \)

However, substituting \(x = -5\) leads to a negative argument, making the logarithm undefined:

• \( \log_3(-5-4) \) is undefined.

Therefore, it's important to consider these validity checks to ensure that each solution is feasible.