Problem 43 - Precalculus

Question

Solve the logarithmic equations exactly. $$\log _{2}(4 x-1)=-3$$

Step-by-Step Solution

Answer

\(x = \frac{9}{32}\)

Step 1: Understand the Equation

We start with the given logarithmic equation: \(\log _{2}(4x-1)=-3\). This equation tells us that the logarithm of \(4x-1\) with base 2 is equal to -3.

Step 2: Convert Logarithmic to Exponential Form

To eliminate the logarithm, we convert the logarithmic equation to its equivalent exponential form. If \(\log_{b}(a) = c\), then the exponential form is \(b^{c} = a\). Thus, \(2^{-3} = 4x - 1\).

Step 3: Simplify the Exponential Equation

Evaluate \(2^{-3}\). Since \(2^{-3} = \frac{1}{2^3} = \frac{1}{8}\), substitute it into the equation: \(\frac{1}{8} = 4x - 1\).

Step 4: Solve for x

Add 1 to both sides to isolate the terms involving \(x\): \(\frac{1}{8} + 1 = 4x\). Express \(1\) as \(\frac{8}{8}\) and add: \(\frac{1}{8} + \frac{8}{8} = \frac{9}{8}\). So, \(\frac{9}{8} = 4x\).

Step 5: Divide to Find x

Divide both sides by 4 to solve for \(x\): \(x = \frac{9}{8} \div 4 = \frac{9}{8} \times \frac{1}{4} = \frac{9}{32}\).

Key Concepts

Exponential Form

When studying logarithmic equations, it is essential to understand the concept of exponential form. This form is a way to represent numbers using a base and an exponent. The exponential expression is generally written as \(b^c = a\), where:
\(b\) is the base,
\(c\) is the exponent,
\(a\) is the result of multiplying the base, \(b\), by itself \(c\) times.
For example, \(2^3 = 8\), which means that 2 is multiplied by itself three times to give the result of 8. Thus, in the exponential form, you can see the direct relationship between the base, exponent, and the result.
Logs are closely related to exponents and serve as the opposite operation. Understanding this concept is vital for solving logarithmic equations, as it allows us to transform equations into a form that is typically easier to solve.

Solving Equations

Solving equations involves finding the value of variables that satisfy the given mathematical statement. In the case of a logarithmic equation, we often aim to isolate the variable by transforming the equation into a more manageable form, preferably by converting it into an exponential form. Using our given logarithmic equation \(\log _{2}(4x-1)=-3\) as an example:

• We convert this into exponential form, transforming it into \(2^{-3} = 4x - 1\).
• This step makes the equation easier to handle by eliminating the logarithm.
• Now, solve the equation by performing operations to isolate the variable \(x\).

By doing this, we can directly solve for \(x\) by adding, subtracting, multiplying, or dividing both sides of the equation, as needed. This process often involves basic arithmetic operations, and once simplified, provides the solution needed to satisfy the original equation.

Conversion between Logarithmic and Exponential Forms

One crucial skill in solving logarithmic equations is the conversion between logarithmic and exponential forms. This conversion helps simplify complex-looking logarithmic equations into a more straightforward exponential form, which is easier to handle and solve.
Here is how the conversion works:

• A logarithmic expression looking like \(\log_b(a) = c\) can be converted to \(b^c = a\) in exponential form.
• Conversely, the exponential form \(b^c = a\) can be translated back into logarithmic form as \(\log_b(a) = c\).

This technique is particularly useful because it allows you to address the problem using skills from elementary algebra, improving your ability to find solutions efficiently. Transforming between these forms is like switching perspectives and gives you a different approach to solve the problem. So, the next time you encounter a logarithmic equation, remember to consider converting it into an exponential form to solve it effectively.