Problem 56 - Intermediate Algebra

Question

Solve for \(x .\) See Example 3. $$ \log _{4} 16=x $$

Step-by-Step Solution

Answer

The solution is \( x = 2 \).

Step 1: Understanding Logarithm Definition

The given problem is to solve the logarithmic equation \( \log_{4} 16 = x \). Recall that the logarithm \( \log_{b} a = c \) is equivalent to saying \( b^c = a \). Here, \( b = 4\), \( a = 16 \), and we need to find \( x \) such that \( 4^x = 16 \).

Step 2: Express 16 as a Power of 4

Our next task is to express 16 using a base of 4. Notice that 16 can be expressed as a power of 4: \( 16 = 4^2 \). This is because when you multiply 4 by itself, you get 16: \( 4 \times 4 = 16 \), hence \( 4^2 = 16 \).

Step 3: Equate and Solve

Now that we know \( 16 = 4^2 \), we can rewrite the equation \( 4^x = 16 \) as \( 4^x = 4^2 \). Since the bases are equal, we can set the exponents equal to each other: \( x = 2 \).

Step 4: Verify the Solution

To ensure our solution is correct, verify by substituting \( x = 2 \) back into the original expression: \( \log_{4} 16 = x \) becomes \( \log_{4} 16 = 2 \) because \( 4^2 = 16 \). This confirms that \( x = 2 \) is indeed the correct solution.

Key Concepts

Logarithm Properties

Logarithms are a fundamental concept in mathematics that have the power to simplify complex problems and solve various equations. A logarithm, like \(\log_{b} a = c\), tells us what power we need to raise the base \(b\) to get the number \(a\). This fundamental idea helps in transforming multiplicative processes into additive ones, making calculations more manageable.

Some essential properties of logarithms include:

• Logarithm of a Product: \( \log_b (MN) = \log_b M + \log_b N \)
• Logarithm of a Quotient: \( \log_b \left(\frac{M}{N}\right) = \log_b M - \log_b N \)
• Logarithm of a Power: \( \log_b (M^p) = p \cdot \log_b M \)
• Change of Base Formula: \( \log_b a = \frac{\log_k a}{\log_k b} \)

These properties allow us to manipulate and break down logarithmic equations with ease. For instance, understanding that \(\log_{4} 16 = x\) based on the property \(b^c = a\) lets us convert it into an exponential form \(4^x = 16\), assisting further in solving the equation.

Exponential Equations

Exponential equations involve a variable in the exponent and can often appear intimidating. However, by converting logarithmic equations to their exponential forms, they become easier to handle. The key is understanding that an equation like \(\log_b a = c\) implies \(b^c = a\). This makes our task to determine the exponent when a number is raised to a power.

When solving exponential equations:

• Express both sides of the equation with the same base, if possible.
• Once the bases are identical, set the exponents equal to each other.
• Solve this simpler algebraic equation to find the variable.

In the problem \(\log_{4} 16 = x\), converting it into \(4^x = 16\) reveals that the equation is manageable by expressing 16 as \(4^2\). This way, matching bases \(4^x = 4^2\) directly gives us the solution \(x = 2\).

Mastering exponential equations is about practice and recognizing patterns, greatly aided by understanding their logarithmic connections.

Base of a Logarithm

The base of a logarithm is a crucial element that dictates the terms of the equation. It is the number that, when raised to a specific power, results in another number. This concept is clearly illustrated in the logarithmic equation \(\log_{4} 16 = x\), where 4 is the base.

Understanding the base helps you:

• Determine the power or exponentiate result needed to solve for \(x\).
• Identify patterns in numbers or expressions, making calculations simpler.

For instance, by recognizing the base 4 and rewriting the number 16 as \(4^2\), students can directly find the exponent \(x\). Knowing common bases and their powers is beneficial as they often appear in various problems, such as:

• Base 10, commonly used in scientific applications (common logarithms).
• Base \(e\) (approximately 2.718), vital in calculus and natural growth models (natural logarithms).

Focusing on the base provides clarity in converting equations and simplifies the overall problem-solving process. Remain observant of base values to find the most straightforward path to solving logarithmic and exponential equations.