Problem 44 - Elementary and Intermediate Algebra

Question

Solve each equation. See Example \(6 .\) $$ \log _{4}(2 x-1)=3 $$

Step-by-Step Solution

Answer

The solution is \( x = 32.5 \).

Step 1: Use the Definition of a Logarithm

Recall that the logarithm equation \( \log_b(a) = c \) can be rewritten in exponential form as \( b^c = a \). Here, \( b = 4 \), \( a = 2x-1 \), and \( c = 3 \). Thus, rewrite the equation as \( 4^3 = 2x - 1 \).

Step 2: Calculate the Power

Compute \( 4^3 \), which means 4 raised to the power of 3. \( 4^3 = 4 \times 4 \times 4 = 64 \). Hence, the equation becomes \( 64 = 2x - 1 \).

Step 3: Solve for x

Start by isolating \( 2x \) on one side of the equation. Add 1 to both sides to get \( 64 + 1 = 2x \). Simplifying the left side gives \( 65 = 2x \).

Step 4: Divide to Find x

To solve for \( x \), divide both sides of the equation by 2. Thus, \( x = \frac{65}{2} = 32.5 \).

Step 5: Verify the Solution

Substitute \( x = 32.5 \) back into the original equation to check if it holds true. Verify by calculating \( 2x - 1 = 2(32.5) - 1 = 65 - 1 = 64 \), and check \( \log_4(64) \). Since \( 4^3 = 64 \), the solution is correct.

Key Concepts

Logarithms

Logarithms are the inverse operation of exponentiation, serving as a tool to determine the power to which a given base must be raised to produce a certain number. In simple terms, if you have an equation of the form \( \log_b(a) = c \), it means that raising the base \( b \) to the power of \( c \) will yield \( a \).

• This inverse relationship is crucial, enabling us to switch between logarithmic and exponential forms easily.
• For instance, if \( \log_4(64) = 3 \), the equivalent exponential form is \( 4^3 = 64 \).

Understanding and manipulating logarithms can simplify complex equations. Remember that the base of a logarithm cannot be a negative number or zero, and the arguments inside logarithms must also be positive numbers to ensure real number solutions. Often, converting logarithmic equations into exponential form makes them easier to solve.

Exponential Equations

Exponential equations typically involve variables in the exponent, making them a bit more complex to handle than linear equations. They are seen in the form \( b^c = a \), where \( b \) is the base and \( c \) is the exponent; such equations are pivotal in modeling growth and decay processes, among others.

• To solve exponential equations like the one in our exercise, you can often utilize logarithms to bring the variable down from the exponent.
• When manipulating these equations, it is crucial to calculate the exponential power carefully to ensure accuracy, like computing \( 4^3 \) in the given problem, which results in 64.

Once the equation is transformed into a simpler arithmetic form, solving for the variable becomes straightforward. Always remember to verify the solution by plugging it back into the original equation to ensure it satisfies the problem conditions.

Problem Solving Techniques

Tackling math problems, especially involving logarithms and exponents, can be simplified using structured problem-solving techniques. These techniques help to systematically break down the problem.

• Step-by-Step Approach: Start by understanding what is being asked. For instance, in logarithmic problems, convert the equation into an exponential form to glean more intuitive insights.
• Calculation and Isolation: Perform necessary calculations like finding exponential values and isolate the variable of interest to determine its value, as shown in the example where \( x = \frac{65}{2} \).
• Verification: Confirm your solution by substituting back into the original equation. This crucial verification step ensures the solution satisfies all initial conditions.

Employ these strategies consistently, and they will bolster your ability to tackle similar mathematical challenges. Remember, practice is key in mastering these techniques.