Problem 44 - Precalculus

Question

Solve the logarithmic equations exactly. $$\log _{4}(5-2 x)=-2$$

Step-by-Step Solution

Answer

The solution is \( x = \frac{79}{32} \).

Step 1: Understand the Logarithmic Equation

The given equation is \( \log_{4}(5-2x) = -2 \). This means we have the logarithm (base 4) of \((5-2x)\) equals \(-2\). Our goal is to express this equation in its exponential form to find the value of \(x\).

Step 2: Convert the Logarithmic Equation to Exponential Form

Since \( \log_{b}(y) = x \) is equivalent to \( b^{x} = y \), we use this property to rewrite the equation: \[ 4^{-2} = 5-2x \].

Step 3: Simplify the Exponential Equation

Calculate \(4^{-2}\), which means taking the reciprocal of \(4^2\):\[ 4^{-2} = \frac{1}{4^2} = \frac{1}{16} \].The equation becomes:\[ \frac{1}{16} = 5 - 2x \].

Step 4: Isolate the Term Involving x

Rearrange the equation to solve for \(x\). Subtract 5 from both sides:\[ \frac{1}{16} - 5 = -2x \].First, express 5 in terms of a common denominator:\[ \frac{1}{16} - \frac{80}{16} = -2x \].The equation becomes:\[ \frac{-79}{16} = -2x \].

Step 5: Solve for x

To solve for \(x\), divide both sides by \(-2\):\[ x = \frac{-79}{16} \times \frac{-1}{2} = \frac{79}{32}. \]

Step 6: Final Check

We need to verify the solution by substituting \(x = \frac{79}{32}\) back into the original equation to ensure it holds true:\[ 5 - 2 \times \frac{79}{32} = 5 - \frac{158}{32} = 5 - \frac{79}{16} = \frac{80}{16} - \frac{79}{16} = \frac{1}{16}. \]Taking the log base 4 of \( \frac{1}{16} \):\[ \log_{4}\left(\frac{1}{16}\right) = -2. \]The solution is verified.

Key Concepts

Understanding Exponential Form

When dealing with logarithmic equations, converting them to exponential form is a key strategy for finding solutions. But what does this mean? An equation of the form \( \log_{b}(y) = x \) tells us that the base \( b \) raised to the power \( x \) equals \( y \). So, converting to exponential form means expressing this relationship as \( b^{x} = y \).

This conversion is helpful because it transforms often complex logarithmic equations into simpler exponential equations easier to solve. In the given example, \( \log_{4}(5-2x) = -2 \) becomes \( 4^{-2} = 5 - 2x \) when converted. By solving \( 4^{-2} \), you simplify your problem to handle traditional algebraic techniques.

If you remember this vital step, tackling logarithmic equations effectively becomes much more straightforward.

Exploring Logarithmic Functions

Logarithmic functions are fascinating yet sometimes confusing concepts. They are essentially the inverse of exponential functions. While an exponential function like \( y = b^{x} \) uses exponents to transform a base, a logarithmic function \( x = \log_{b}(y) \) "undoes" this by telling us what exponent on the base gets us back to \( y \).

Logarithms have bases, just like exponential functions. In your exercise, the base is 4. This base is crucial because it defines the scale or "growth" rate within the function. Understanding log properties enables further manipulation and solving of complex equations.

The primary property you'll often rely on is that if \( \log_{b}(y) = x \), then it implies \( b^{x} = y \). This property makes conversions and solving possible, so mastering these functions will empower your mathematical skills.

Strategies for Solving Equations

When solving equations, particularly logarithmic ones, structured approaches offer clarity and efficiency. The fundamental strategy involves converting the given logarithmic equation into an exponential form and solving it using algebraic methods. Let's walk through some general tips.

• **Identify the Logarithmic Equation:** Check its structure and base to understand the given equation fully.
• **Convert to Exponential Form:** Use the equation \( \log_{b}(y) = x \) as \( b^{x} = y \) to simplify the equation.
• **Simplify Using Algebraic Techniques:** After converting, solve the equation by isolating the variable and simplifying, as with any standard algebra problem.
• **Check Your Work:** Substitute your solution back to ensure it's valid, which can avoid simple errors and assure you the solution is correct.

The initial complexity of logarithmic equations often becomes manageable once these steps are systematically applied.