Problem 32 - Algebra for College Students

Question

Solve each logarithmic equation and express irrational solutions in lowest radical form. $$ \log x^{2}=(\log x)^{2} $$

Step-by-Step Solution

Answer

The solutions are \( x = 1 \) and \( x = 100 \).

Step 1: Understand the Problem

We have the equation \( \log x^2 = (\log x)^2 \). Our goal is to find the value(s) of \( x \) that satisfy this equation.

Step 2: Simplify the Left Side

Apply the power rule for logarithms on the left side: \( \log x^2 = 2 \log x \). This simplifies the equation to \( 2 \log x = (\log x)^2 \).

Step 3: Set the Equation to Zero

Rearrange the equation to have zero on one side by subtracting \( 2 \log x \): \( (\log x)^2 - 2\log x = 0 \).

Step 4: Factor the Equation

Notice that this can be factored as a quadratic in terms of \( \log x = y \): \( y^2 - 2y = 0 \). Factor to get \( y(y - 2) = 0 \).

Step 5: Solve for \( y \)

Set each factor to zero: \( y = 0 \) or \( y - 2 = 0 \). This gives us the solutions \( y = 0 \) and \( y = 2 \).

Step 6: Interpret in Terms of \( x \)

Recall \( y = \log x \). Substitute \( y = 0 \) to get \( \log x = 0 \), so \( x = 10^0 = 1 \). Substitute \( y = 2 \) to get \( \log x = 2 \), so \( x = 10^2 = 100 \).

Step 7: Verify Conditions

Confirm that both solutions \( x = 1 \) and \( x = 100 \) satisfy the original equation: both \( 2 \log 1 = (\log 1)^2 = 0 \) and \( 2 \log 100 = (\log 100)^2 = 4 \) are true.

Key Concepts

Power Rule for Logarithms

The power rule for logarithms is a handy tool that simplifies equations involving logarithms. This rule states that, for any positive number base and any real exponent, the logarithm of a number raised to an exponent is equal to the exponent times the logarithm of the number.
This can be written as:

• \( \log_b(a^c) = c \cdot \log_b(a) \)

In the original exercise, we applied this rule to the left side of the equation \( \log x^2 \) which transformed it into \( 2 \log x \).
Why is this useful? By applying the rule, we simplify the expression, making it easier to solve the equation. Simplification is a crucial step in problem-solving, allowing us to manipulate terms and pave the way for further techniques to be applied.

Quadratic Equations in Logarithms

When equations involve a variable raised to a power, they often take the form of quadratic equations. In our context, the equation evolved into a quadratic because of its form \( (\log x)^2 - 2\log x = 0 \).
Here, by setting \( \log x = y \), we convert the problem into a more familiar form \( y^2 - 2y = 0 \), which is a standard quadratic. Understanding this transformation is essential because it allows us to utilize various quadratic techniques like factoring.
Quadratics play a pivotal role in mathematics, and recognizing them hidden within logarithmic equations can make solutions straightforward and logical. Once in quadratic form, we can apply techniques we've learned in basic algebra, opening the door for solving seemingly more complex logarithmic equations.

Factorization Techniques

Factorization transforms an equation into a product of simpler expressions, which can simplify finding its solutions. For quadratic equations, especially the type \( y^2 - 2y = 0 \), factorization reveals the solutions easily:

• We can express this as \( y(y - 2) = 0 \).
• If the product of two factors equals zero, then at least one of the factors must be zero, hence providing potential solutions.

In our context, setting each factor to zero provided us with \( y = 0 \) and \( y = 2 \). These were then used to find corresponding values for \( x \) using \( \log x = y \). Factorization simplifies the process of solving polynomial equations, making it an invaluable method in solving different mathematical problems.

Solution Verification

Verification is a critical step in solving equations. It's essential to check that the solutions found satisfy the original equation.
For our problem, after finding that \( x = 1 \) and \( x = 100 \), we substitute back into the original logarithmic equation \( \log x^2 = (\log x)^2 \) to see if they hold true:

• For \( x = 1 \): \( 2\log 1 = (\log 1)^2 = 0 \), which is true.
• For \( x = 100 \): \( 2\log 100 = (\log 100)^2 = 4 \), which also holds true.

Verification ensures accuracy and reinforces confidence in the solutions obtained. It is always a good practice to confirm that the solutions fit within the context of the problem, ensuring no steps have been missed or miscalculated.