Problem 67 - Algebra and Trigonometry

Question

Solve each logarithmic equation. Be sure to reject any value of \(x\) that is not in the domain of the original logarithmic expressions. Give the exact answer. Then, where necessary, use a calculator to obtain a decimal approximation, correct to two decimal places, for the solution. $$ \log _{5} x+\log _{5}(4 x-1)=1 $$

Step-by-Step Solution

Answer

The solution to the logarithmic equation \(\log _{5} x+\log _{5}(4 x-1)=1\) is \(x=1.28\), approximately. The other solution does not adhere to the domain of the original logarithmic expression and is thus rejected.

Step 1: Apply the properties of logarithms

Apply the property of logarithms \(\log _{b} a+\log _{b} c = \log _{b} ac\) to write the two logarithmic expressions on the left as a single logarithm. In this case, it yields: \(\log _{5} (x(4x-1)) = \log _{5} (4x^2-x) = 1\).

Step 2: Convert the logarithmic equation to exponential form

The equation from step 1, \(\log _{5} (4x^2 - x) = 1\), is in logarithmic form. Convert this to exponential form using the property \(b^y = a\) for the logarithm \(\log_{b} a = y\). Thus, the equation becomes: \(5^1 = 4x^2 - x\), which simplifies to: \(5 = 4x^2 - x\).

Step 3: Rearrange the equation to quadratic form

The equation from step 2 can be rearranged to form a quadratic equation. Doing so yields: \(4x^2 - x - 5 = 0\).

Step 4: Solve the quadratic equation

Solve the quadratic equation \(4x^2 - x - 5 =0\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here a = 4, b = -1,and c = -5. Therefore, the solutions of this quadratic equation are: \(x = \frac{1 + \sqrt{1 + 80}}{8}\) or \(x = \frac{1 - \sqrt{81}}{8}\).

Step 5: Check for values within the domain

Remember that for \(\log _{a} x\), \(x > 0\). Therefore, any negative solutions to the quadratic equation will not be within the domain of the original logarithmic equation and must be rejected. In this case, only \(x = \frac{1 + \sqrt{1 + 80}}{8}\) is valid as the other root is not greater than 0.

Step 6: Convert to decimal

Following the given instruction, convert the final solution to a decimal approximation correct to two decimal places. So, the decimal value of \(x= \frac{1 + \sqrt{81}}{8}\) is approximately 1.28.

Key Concepts

Logarithmic Properties

Logarithmic properties are essential tools for simplifying and solving equations involving logarithms. A primary feature is the product rule:

• Product Rule: \( \log_{b} a + \log_{b} c = \log_{b} (ac) \)

This rule combines two logarithmic terms with the same base into a single logarithm. It simplifies expressions and allows us to isolate the variable of interest.
In the given exercise, the product rule combines \( \log_{5} x \) and \( \log_{5} (4x-1) \) to form \( \log_{5} (4x^2-x) \).
This transformation is crucial because it sets up the equation to be more easily converted and solved.

Domain of Logarithmic Functions

Understanding the domain of logarithmic functions is crucial for solving logarithmic equations correctly. The domain of a logarithmic function \( \log_{a} x \) is all positive real numbers:

• The argument, \( x \), must be greater than zero: \( x > 0 \)

This condition stems from the fact that you cannot take the logarithm of a non-positive number.
In our exercise, we had an expression \( \log_{5} (4x^2 - x) \). We must verify that all potential solutions satisfy \( 4x^2 - x > 0 \). Only the positive solution from the quadratic equation is valid.
Thus, applying this check helps ensure the solution makes sense in real-world contexts.

Quadratic Formula

The quadratic formula is a powerful method for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). It provides solutions as:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

This formula helps find solutions by considering coefficients \( a, b, \) and \( c \) from the equation.
In the exercise, the equation \( 4x^2 - x - 5 = 0 \) was solved using the quadratic formula. By identifying \( a = 4, b = -1, \) and \( c = -5 \), the formula efficiently provided potential solutions for \( x \).
Checking the quadratic discriminant \( b^2 - 4ac > 0 \) showed that two real solutions exist, which were then checked against the logarithmic constraints.

Decimal Approximation

Decimal approximation helps in understanding results pragmatically, especially when exact solutions are complex. Once we get the exact solution:

• Convert it to a decimal form using a calculator.
• Round it to the required precision, typically two decimal places.

In the exercise, the solution \( x = \frac{1 + \sqrt{81}}{8} \) was converted to 1.28. This makes the result easier to interpret and use in further applications or comparisons.
Decimal approximation finds frequent use not only in logarithmic equations but also across various mathematical problems to present comprehensible results.