Problem 15 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

Solve for \(x\) : \(\log _{3}(\sqrt{x}+|\sqrt{x}-1|)^{2}=\log _{3}(4 \sqrt{3}-3+4|\sqrt{x}-1|)\)

Step-by-Step Solution

Answer

Therefore, the possible solutions for \(x\) are the ones obtained from solving the two separate equations in step 3 and step 4. However, be sure to verify the solutions by substituting them back into the initial equation.

Step 1: Simplify the equation

As the bases of the logs are both 3, we can simplify the equation to: \(\sqrt{x}+|\sqrt{x}-1|^{2} = 4\sqrt{3} - 3 + 4|\sqrt{x}-1|\)

Step 2: Split the equation based on the absolute value

The absolute value divides the equation into two separate parts, for \(\sqrt{x} - 1 \geq 0\) and for \(\sqrt{x} - 1 < 0\).

Step 3: Solve the equation where \(\sqrt{x} - 1 \geq 0\)

Let's solve the equation for \(\sqrt{x} - 1 \geq 0\):\((2\sqrt{x})^2 = 4\sqrt{3} - 3 + 4\sqrt{x}\) - From here, you get a quadratic equation that you can solve for \(\sqrt{x}\), and finally square the result to find \(x\).

Step 4: Solve the equation where \(\sqrt{x} - 1 < 0\)

Let's solve the equation for \(\sqrt{x} - 1 < 0\): \((2\sqrt{x} - 2)^2 = 4\sqrt{3} - 3 + 4(1 - \sqrt{x})\) - Follow the same steps as in step 3 to find the another set of solutions for \(x\).

Step 5: Combine the solutions

Combine the solutions from step 3 and step 4. Make sure to discard any values that do not meet the original conditions of the equation.

Key Concepts

Absolute Value

In mathematics, the absolute value of a number is its non-negative value without regard to its sign. Absolute value is helpful where we are concerned with the distance of a number from zero on a number line, irrespective of direction. It's denoted by vertical bars: \(|a|\).

• The absolute value of a positive number is the number itself.
• The absolute value of a negative number is its positive counterpart.
• The absolute value of zero is simply 0.

For example, \(|-5| = 5\) and \(|3| = 3\). Absolute values are crucial when dealing with expressions that must always yield non-negative outcomes.
In the exercise, \( |\sqrt{x} - 1|\) breaks the problem into two cases. We must consider both situations: when \(|\sqrt{x} - 1|\) is non-negative and when it's negative. This ensures all solutions respect the behavior of absolute values and provide us a broad picture for solving the equation.

Quadratic Equations

A quadratic equation is a polynomial equation of degree 2. Its standard form is:\[ax^2 + bx + c = 0\]where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). Quadratic equations can have two real roots, one real root (a repeated root), or two complex roots.There are several techniques to solve a quadratic equation:

• Factoring, when the equation can be expressed as a product of its factors.
• Completing the square, which involves turning the equation into a perfect square trinomial.
• The quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

In the exercise, quadratic equations arise after we simplify and rearrange parts of the solved expressions like \((2\sqrt{x})^2\). Solving these quadratic equations reveals potential solutions for \(\sqrt{x}\)\), which in turn allow finding results for \(x\). Remember: quadratics may have multiple solutions, so each root needs verification against initial conditions.

Logarithmic Functions

Logarithmic functions are expressions of the form \(\log_b(y) = x\), meaning \(b^x = y\). They serve as the inverses of exponential functions and are vital in solving equations involving exponentiated terms.Some key attributes of logarithms include:

• The base \(b\) of a logarithmic function is always a positive number.
• The function is defined only for positive \(y\) values.
• Common logarithms (base 10) and natural logarithms (base \(e\)) are particular types.

In our problem, the logs have a common base of 3, allowing us to equate their arguments directly once we simplify the expression inside each log function: \(\log_3(f(x)) = \log_3(g(x))\Rightarrow f(x) = g(x)\). This property helps in transforming complex expressions into simpler ones, ultimately leading to polynomial forms we can solve more easily.
Logarithmic functions are not only integral to calculus but also appear frequently in real-world situations where variables change multiplicatively—such as in finance and science.