Problem 68 - Introductory and Intermediate Algebra for College Students 4th

Question

Solve each logarithmic equation. Be sure to reject any value of \(x\) that is not in the domain of the original logarithmic expressions. Give the exact answer. Then, where necessary, use a calculator to obtain a decimal approximation, correct to two decimal places, for the solution. $$\log _{2}(x-1)+\log _{2}(x+1)=3$$

Step-by-Step Solution

Answer

The exact answer is \(x = 3\).

Step 1: Combine the Logarithmic Terms

Using the property of logarithms that states \(\log _a(bc) = \log _a(b) + \log _a(c)\), the two logs on the left side of the equation can be combined into a single log term:\n\(\log _2(x-1)(x+1) = 3\)

Step 2: Convert to Exponential Form

To eliminate the logarithm, convert the equation from logarithmic form to exponential form. Use the definition of logarithms, which states that \(\log_b{a}=c\) is equivalent to \(b^c = a\). So the equation becomes:\n\(2^3 = (x-1)(x+1)\)

Step 3: Solve for x

Simplify and solve for \(x\):\n\(8 = x^2 - 1\), \n\(x^2 = 9\), and thus \n\(x = \pm \sqrt{9} = \pm 3\)

Step 4: Check the Solutions

Checking that \(x = 3\) and \(x = -3\) are in the domain of the original logarithmic function: \nFor \(x = 3\), both \(3-1\) and \(3+1\) are positive, thus \(x = 3\) is a valid solution. \nFor \(x = -3\), \(-3-1\) is negative, thus \(x = -3\) is not valid since it's not possible to take a logarithm of a negative number.

Key Concepts

Logarithmic Properties

Understanding logarithmic properties is essential for solving logarithmic equations. Logarithmic functions basically reverse the action of exponentiation. One of the key properties we can utilize is the product rule, which states that the logarithm of a product is the sum of the logarithms of its factors: \[ \log_{b}(mn) = \log_{b}(m) + \log_{b}(n) \]
In the exercise, this property is applied to combine two logarithmic expressions into one. This simplification is a vital step because it allows us to transform complex logarithmic equations into a more manageable form.

Combining Logarithms

Another important property is the power rule, which relates powers inside a logarithm to multiplication outside: \[ \log_{b}(m^n) = n\cdot\log_{b}(m) \]
This property wasn't directly used in the exercise provided, but it's another powerful tool in the logarithmic toolkit which could come in handy for other similar problems.

Exponential Form

Transitioning from logarithmic form to exponential form is a fundamental step in solving logarithmic equations. This process utilizes the definition of a logarithm, which can be expressed as: \[ \log_{b}(a) = c \quad \Rightarrow \quad b^c = a \]
Using this conversion allows us to move from a world of logarithms to the more familiar territory of exponential equations. In context, for the given exercise, converting \( \log_{2}(x-1)(x+1) = 3 \) into exponential form simplifies it to \( 2^3 = (x-1)(x+1) \) which is crucial for finding the potential values of 'x'.

From Logarithms to Exponents

This step takes the base 'b' of the logarithm, raises it to the 'c' on the other side of the equation, and sets it equal to 'a'. It flips the equation on its head, moving us towards a quadratic equation that can be solved using algebra.

Domain of Logarithmic Functions

Logarithmic functions are not defined for zero or negative numbers. This restriction is because we can only take logarithms of positive numbers. Hence, the domain of a logarithmic function includes all positive real numbers. In solving logarithmic equations, any value for 'x' that produces a negative number or zero within the logarithmic function must be rejected, since it falls outside the domain.

Checking the Validity

For example, in the given exercise, after solving the transformed equation, we obtain two possible solutions: \(x = 3 \) and \(x = -3\). However, we must check these solutions against the original logarithmic expressions. \(x = -3\) results in a negative number inside one of the logarithmic expressions, which violates the domain restrictions of logarithmic functions. Consequently, \(x = -3\) is not a valid solution, whereas \(x = 3\) satisfies the domain constraints comfortably since both \(3-1\) and \(3+1\) are positive numbers.