Problem 10 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

Solve for \(x\) : \(\frac{1}{2} \log _{5}(x+5)+\log _{5}(\sqrt{x-3})=\frac{1}{2} \log _{5}(2 x+1)\)

Step-by-Step Solution

Answer

The solution to the equation is \(x = 8\).

Step 1: Simplify using properties of logarithms

To simplify this equation, the properties of logarithms are used: \(\log_b(mn) = \log_b(m) + \log_b(n)\) and \(\log_b(m^n) = n \log_b(m)\). First, apply the power rule to get:\(\frac{1}{2} \log _{5}(x+5)+\log _{5}((x-3)^{1/2})=\frac{1}{2} \log _{5}(2x+1)\) Then, apply the product rule to combine the logarithms on the left side:\(\log _{5}((x+5)^{1/2} * (x-3)^{1/2})=\frac{1}{2} \log _{5}(2x+1)\)

Step 2: Convert from logarithmic to exponential form

Now, convert the equation from logarithmic form to exponential form. The general rule to convert from logarithm to exponent is: if \(\log_a(b) = c\), then \(a^c = b\).Applying this rule yields:\((5^{1/2} * (2x+1)^{1/2}) = (x+5)^{1/2} * (x-3)^{1/2}\)

Step 3: Solve for \(x\)

After squaring both sides to remove the square root, and simplifying, you get the quadratic equation \(x^2 - 7x = 8\). Solving for \(x\) by using the quadratic formula \(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), the roots of the quadratic equation are obtained. Considering the domain of the original logarithmic equation (both arguments of the logs must be positive), one can verify which root(s) are valid.

Key Concepts

Properties of Logarithms

Logarithms have several key properties that make them extremely useful for solving various types of equations. One essential property is the **product rule**, which states that the logarithm of a product is the sum of the logarithms of the factors. Mathematically, this is written as:

• \(\log_b(mn) = \log_b(m) + \log_b(n)\)

Another vital property is the **power rule**, which allows you to remove an exponent within a logarithmic function by multiplying the logarithm by the exponent. It is expressed as:

• \(\log_b(m^n) = n \cdot \log_b(m)\)

These properties simplify comparisons and calculations involving logarithmic equations. Understanding them is critical for effectively manipulating terms, as seen in the original exercise.

Exponential Form Conversion

Converting a logarithmic equation to its exponential form is a powerful technique used to simplify and solve for the unknown variable. The general rule to perform this conversion is:

• If \(\log_a(b) = c\), then it can be rewritten in exponential form as \(a^c = b\).

This conversion unveils a more direct relationship between the base, exponent, and result, making it easier to manipulate. In the example provided, rewriting in exponential form helped to equate both sides of the equation clearly, ultimately leading to a solvable quadratic equation.

Quadratic Equation

Quadratic equations are polynomial equations of the form \(ax^2 + bx + c = 0\). These equations are essential in mathematics and occur often when solving logarithmic equations converted to polynomial form. In the step-by-step solution, the equation \(x^2 - 7x = 8\) appears. One standard method to solve this quadratic equation is by using the **quadratic formula**, which is given by:

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

Understanding how to apply this formula is crucial, as it finds the roots of the quadratic equation, which are the potential solutions for \(x\). Determining whether all potential solutions are valid necessitates a check against any conditions imposed by the original problem, such as the domain of a logarithmic function.

Logarithmic Function Domain

The domain of a logarithmic function is the set of all possible input values (arguments) for which the function is defined. Since the logarithm \(\log_b(x)\) is only defined for positive arguments (\(x > 0\)), this restriction influences the solutions to logarithmic equations. In our exercise, both \(x + 5\) and \(\sqrt{x - 3}\) must be positive for the logs to be valid. This implies conditions:

• x + 5 > 0, which simplifies to \(x > -5\)
• sqrt{x - 3} implies \(x - 3 > 0\), simplifying to \(x > 3\)

Therefore, any solution to the equation must satisfy \(x > 3\). This ensures that both the square root and logarithmic parts of the equation remain defined, making a domain check a vital part of solving logarithmic equations accurately.