Problem 51 - Elementary and Intermediate Algebra

Question

Solve for \(x\). See Example 3 . $$ \log _{x} 81=2 $$

Step-by-Step Solution

Answer

The solution for \( x \) is 9.

Step 1: Identify the Equation

The given logarithmic equation is \( \log_{x} 81 = 2 \). The equation uses logarithms, which have the general form \( \log_b y = c \). This means \( b^c = y \).

Step 2: Transform the Logarithmic Equation

By using the properties of logarithms, transform \( \log_{x} 81 = 2 \) into its exponential form. The exponential form of the logarithm \( \log_{x} 81 = 2 \) is \( x^2 = 81 \).

Step 3: Solve the Exponential Equation

Now solve the equation \( x^2 = 81 \) for \( x \). Recognize that 81 is a perfect square: \( 81 = 9^2 \). Thus, \( x^2 = 9^2 \).

Step 4: Find the Value of x

Since \( x^2 = 9^2 \), we can take the square root of both sides to find \( x \). This gives \( x = 9 \) or \( x = -9 \).

Step 5: Confirm Suitable Solutions

Since the base of a logarithm must be positive and not equal to 1, check the possible solutions: \( x = 9 \) satisfies \( x > 0 \). However, \( x = -9 \) does not, as a logarithmic base cannot be negative.

Key Concepts

Properties of Logarithms

Logarithms hold a curious array of properties that make them essential tools in mathematics. Essentially, a logarithm is the inverse of an exponentiation and is a powerful way to express exponential relationships. If we take an equation like \( \log_b y = c \), this means the base \( b \) raised to the power \( c \) results in \( y \), hence \( b^c = y \). This is called the exponential form. Logarithms help in solving equations especially when the exponential form is given but the base or the exponent is unknown. They simplify expressions and help convert multiplication into addition, leveraging their unique properties.
These properties include:

• \( \log_b (xy) = \log_b x + \log_b y \)
• \( \log_b \left(\frac{x}{y}\right) = \log_b x - \log_b y \)
• \( \log_b x^c = c \cdot \log_b x \)

In our exercise, the property used transforms the logarithmic equation into an exponential equation: from \( \log_x 81 = 2 \) to \( x^2 = 81 \). This shows how logarithms effectively simplify the problem, setting the stage for further solving.

Exponential Equations

An exponential equation is one where the variable appears in the exponent. Solving exponential equations often requires transforming them via logarithms or by equating the exponents after both sides of the equation have the same base. For example, in the exercise, once we convert \( \log_x 81 = 2 \) to \( x^2 = 81 \), we have an exponential equation with \( x \) as the base.

• To solve it, we need recognition of perfect squares or can use logarithms to find the variable.
• In our example, \( x^2 = 81 \) directly implies \( x = 9 \) or \( x = -9 \) since 81 is \( 9^2 \).

Through this method, we harness the simplicity of powers and roots. Remember, the key here is reducing the complexity through transformation, such as the conversion from logarithmic to exponential form, making the solution path clearer and more straightforward.

Solving for x

In math problems like our exercise, the ultimate goal is often to solve for \( x \). Solving for \( x \) means finding the value of the variable that makes the equation true. In our scenario, we transformed the logarithmic equation \( \log_x 81 = 2 \) into \( x^2 = 81 \) for easier solving.

• Recognizing that \( x^2 = 81 \) is a perfect square, we find that \( x = 9 \) or \( x = -9 \).
• After solving, it’s crucial to validate potential solutions. As logarithmic bases must be positive and not equal to 1, correspondingly:

\( x = 9 \) is valid while \( x = -9 \) isn't valid since negative numbers cannot serve as bases for logarithmic functions. Hence, the solution, subjected to these constraints, is \( x = 9 \). This careful process of solving and verifying ensures accuracy and complete understanding of the problem at hand.