Problem 63 - Algebra 2

Question

PREREQUISITE SKILL Solve each equation or inequality. Check your solutions. $$ \log _{10} 2^{x}=\log _{10} 32 $$

Step-by-Step Solution

Answer

x = 5

Step 1: Understand the Equation

The given equation is \( \log _{10} 2^{x} = \log _{10} 32 \). This equation states that the logarithms with base 10 of two terms are equal. Remember, if two logarithms with the same base are equal, then their arguments must also be equal.

Step 2: Apply the Property of Logarithms

Since \( \log_{10} a = \log_{10} b \) implies \( a = b \), we can set the arguments of the logarithms equal to each other: \( 2^x = 32 \). This simplifies the equation to \( 2^x = 32 \).

Step 3: Express 32 as a Power of 2

Note that 32 can be expressed as a power of 2. By calculating, we find that \( 32 = 2^5 \). Thus, we can rewrite the equation as \( 2^x = 2^5 \).

Step 4: Equate the Exponents

Once the bases are the same, you can equate the exponents. Since \( 2^x = 2^5 \), we have \( x = 5 \).

Step 5: Check the Solution

Verify the solution by substituting \( x = 5 \) back into the original equation. Calculate \( \log_{10} 2^{5} \) and \( \log_{10} 32 \). Both expressions equal, confirming our solution. Thus, the solution \( x = 5 \) is correct.

Key Concepts

Logarithmic Properties

Logarithms are powerful tools in math for dealing with exponential relationships. At their core, logarithms help us unravel equations that involve exponents by turning multiplication into addition. This property originates from the definition of a logarithm: if \( b^y = x \), then \( \log_b(x) = y \). Logarithms with the same base follow specific rules that make them consistent tools for solving equations.

Key properties include:

• Product Property: \( \log_b(MN) = \log_b(M) + \log_b(N) \)
• Quotient Property: \( \log_b\left(\frac{M}{N}\right) = \log_b(M) - \log_b(N) \)
• Power Property: \( \log_b(M^n) = n \log_b(M) \)
• Equality Property: If \( \log_b(A) = \log_b(B) \), then \( A = B \)

These properties make solving equations with logs more straightforward. They simplify complex products or quotients into easier equations, break down powers, and balance equations efficiently. In the original exercise, the equality property was essential, where equal logarithmic values implied equal arguments.

Exponential Equations

Exponential equations involve variables in the exponent. Such equations take the form \( b^x = c \), where the goal is usually to find the unknown \( x \). We often use logarithms to solve these equations. Logarithms can efficiently "undo" an exponent by converting the equation into a manageable form.

In the exercise, once we had \( 2^x = 32 \), recognizing that 32 was a power of 2 helped immensely. Understanding common powers can speed up problem-solving:

• \( 2^2 = 4 \)
• \( 2^3 = 8 \)
• \( 2^4 = 16 \)
• \( 2^5 = 32 \)

This knowledge allowed us to transform the equation to \( 2^x = 2^5 \), leading directly to \( x = 5 \) by equating exponents. Many exponential equations work by recognizing similar bases, then focusing on solving what's left.

Problem-Solving Skills

Solving mathematical problems like the one in the exercise requires a strategy and analytical thinking. The key is breaking down the problem into smaller, manageable steps.

Start by thoroughly understanding the equation. Recognize what is being asked and identify known information, such as the base of a logarithm or a recognizable exponent pattern.

Utilizing key logarithmic properties can simplify complex tasks. In this case, using the property that equates arguments when logarithmic statements are equal was crucial. Turn more difficult components into simpler expressions whenever possible.

• A big step is to recognize patterns, like seeing powers of a base or standard logarithmic transformations.
• Verify solutions at the end, ensuring everything aligns with the original equation.

Each problem is unique, but honing these skills enables better use of the mathematical toolbox, ensuring efficient and accurate solutions.