Question

Solve the logarithmic equation algebraically. Approximate the result to three decimal places.\(\log _{4} x-\log _{4}(x-1)=\frac{1}{2}\)

Short answer

The solution to the logarithmic equation is \(x = 2.000\)

Step-by-step solution

Simplify using logarithm properties

Using the logarithmic properties, \(\log _{b} a-\log _{b} c=\log _{b}\left(\frac{a}{c}\right)\). This allows us to write: \(\log _{4} \left(\frac{x}{x-1}\right)=\frac{1}{2}\)

Convert to an exponential equation

The equation \(\log _{b} a=c\) is equivalent to \(b^{c}=a\). Thus we can convert our logarithm equation to an exponential form: \(4^{\frac{1}{2}} = \frac{x}{x-1}\)

Solve for X

From previous step 4^(1/2) is equivalent to 2, this gives us \(2 = \frac{x}{x-1}\). Next, clear the fraction by multiplying both sides by \(x - 1\). This gives \(2(x - 1) = x\) or \(2x - 2 = x\). Solving this equation gives \(x = 2\) as a solution.

Check the Solution

Test the solution in the original equation: \(\log_{4}2 - \log_{4}(2-1) = \frac{1}{2}\). By executing this we see that it results in \(\frac{1}{2} = \frac{1}{2}\) confirming that x=2 is the correct solution.

Key concepts

Understanding Logarithmic Properties

Logarithmic properties are essential when solving logarithmic equations. One of the fundamental properties is the difference of logs property: \(\log_b a - \log_b c = \log_b \left( \frac{a}{c} \right)\). This property helps in simplifying expressions involving multiple logarithms of the same base.

In the given problem \(\log_4 x - \log_4 (x-1) = \frac{1}{2}\), we can apply this property to combine the logarithms into one expression: \(\log_4 \left( \frac{x}{x-1} \right) = \frac{1}{2}\). By simplifying the equation in this manner, we prepare it for further transformation, enhancing the ease of solving the equation algebraically.

Mastering these properties allows us to manipulate and solve logarithmic equations effectively. The key is to simplify, leverage these properties, and transform them for easier handling.

Converting to Exponential Form

To solve logarithmic equations, converting them to exponential form can be very helpful. When we have an equation in the form of \(\log_b a = c\), it can be rewritten in exponential form as \(b^c = a\). This transformation lets us solve for the variable more straightforwardly.

In our equation \(\log_4 \left( \frac{x}{x-1} \right) = \frac{1}{2}\), we apply this property to convert the equation into \(4^{\frac{1}{2}} = \frac{x}{x-1}\). The exponential form is often easier to understand and solve because it turns the problem into a different type of algebraic problem, focusing on powers instead of logarithms.

Exponential form conversions are a fundamental step in making complex logarithmic equations more tangible and solvable.

Achieving an Algebraic Solution

Once the equation is in exponential form, solving for the unknown variable requires traditional algebraic techniques. In our example, we have \(4^{\frac{1}{2}} = \frac{x}{x-1}\). Simplifying \(4^{\frac{1}{2}}\) gives us 2, since square roots of numbers are the inverses of squaring.

Now we have \(2 = \frac{x}{x-1}\). To remove the fraction, multiply both sides by \(x - 1\), resulting in: \(2(x - 1) = x\). Distribute and simplify it to \(2x - 2 = x\).

Solve for \(x\) by moving all terms involving \(x\) to one side: \(2x - x = 2\), which gives \(x = 2\). Solving the equations algebraically follows simple logic by isolating the variable and refining the equation step by step.

This approach highlights how logarithmic problems often transition into solving linear or quadratic equations for the unknown.

Approximating Solutions

Logarithmic equations sometimes require approximation to check or validate solutions. However, in this exercise, the solution \(x=2\) is exact and no further decimal approximation is needed. The idea of approximation is significant in scenarios where solutions cannot be expressed as neat integers or straightforward fractions.

Approximation usually comes into play when dealing with transcendental equations or when determining values close to the solution due to rounding. Techniques such as estimating decimals or using logarithmic tables/calculators help with determining these values accurately within specified precision.

It’s crucial for students to recognize when approximations benefit them, especially in real-world applications where precision and accuracy balance practicality.