Problem 96 - Precalculus : Building Concepts and Connections

Question

Solve using any method, and eliminate extraneous solutions. $$\log _{5}|x-2|=2$$

Step-by-Step Solution

Answer

The solutions to the equation are \( x=27 \) and \( x=-23 \).

Step 1: Remove the log by conversion

Remembering that log base \( a \) of \( b \) equal to \( c \) implies that \( b = a^c \), we can convert the logarithmic equation to an exponential equation. Thus, the equation \( \log _{5}|x-2|=2 \) becomes \( |x-2|=5^2 \).

Step 2: Remove the absolute value

Given that \( |a|=b \) implies \( a=b \) or \( a=-b \), the equation \( |x-2|=5^2 \) becomes \( x-2=5^2 \) or \( x-2=-5^2 \). So, we have two equations to solve: \( x-2=25 \) and \( x-2=-25 \).

Step 3: Find the roots

By isolating \( x \) in both equations we gain two roots. For the equation \( x-2=25 \) we have \( x=25+2=27 \). For the equation \( x-2=-25 \) we have \( x=-25+2=-23 \).

Step 4: Check for extraneous solutions

Substituting both obtained roots back into the original equation to eliminate the extraneous. For \( x=27 \), we have \( \log _{5}|27-2|= \log _{5}25 =2 \) which is true. For \( x=-23 \), we have \( \log _{5}|-23-2|= \log _{5}25 =2 \) which is also true. Thus, there are no extraneous solutions.

Key Concepts

Logarithmic and exponential conversion

When we encounter a logarithmic equation, it can often be simplifie by employing the concept of logarithmic and exponential conversion. A logarithmic equation is an equation that includes a logarithm of an unknown variable, typically presented in the form \( \log_{a}x = c \), where \( a \), \( x \), and \( c \) represent known values. To convert this logarithmic equation into an exponential form, we use the fundamental property of logarithms: if \(\log_{a}b = c \), then \(a^c = b \).
The conversion process enables us to solve for the unknown variable more directly. For instance, in the given exercise, the logarithmic equation \(\log_{5}|x-2|=2 \) was converted to its exponential equivalent \(|x-2|=5^2 \). This simplifies the equation significantly, transforming it into a simple absolute value equation that we can manage using algebraic techniques.

Absolute value in equations

The absolute value of a number is its distance from zero on the number line, regardless of direction, which means it is always non-negative. An absolute value equation involves an expression within absolute value bars, for example, \(|x|=a \).
To solve an absolute value equation, we recognize that there are two scenarios to consider: when the inside of the absolute value is positive and when it's negative. Hence, the equation \(|x|=a \) branches into two separate equations: \(x=a \) and \(x=-a \).
Applying this concept to our exercise, \( |x-2|=5^2 \) led to two possibilities, \( x-2=25 \) and \( x-2=-25 \), each yielding distinct solutions. It's crucial to consider both cases, as they correspond to different points on the number line that have the same distance (25, in this case) from the origin.

Extraneous solutions in logarithms

Extraneous solutions are false roots that emerge from the process of solving equations, which do not satisfy the original equation but may appear valid through the algebraic manipulations. In the context of logarithms, extraneous solutions may occur when we remove the logarithm to simplify the equation, particularly when dealing with absolute value expressions or squaring both sides of an equation.
To identify and eliminate extraneous solutions, any potential solution must be plugged back into the original equation to check its validity. In the provided exercise, after finding our potential solutions \(x=27 \) and \(x=-23\), we substituted both into the original logarithmic equation. Since both satisfied the equation, neither of them were extraneous. This verification step is essential, as assuming all algebraically derived solutions are correct without checking can lead to errors in the final answer.