Problem 59 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

\(\log _{4}\left(x^{2}+x\right)-\log _{4}(x+1)=2\)

Step-by-Step Solution

Answer

The solutions to the equation are \(-2\) and \(8\), but \(-2\) is an extraneous solution, so \(x = 8\) is the solution to the original equation.

Step 1: Use the quotient rule for logarithms

Since there is a subtraction between two logarithms with same base, we can use the properties of logarithmic function to combine them into a single logarithm. The quotient rule of logarithms states that \(\log_b{A} - \log_b{B} = \log_b{(A/B)}\). Therefore, the logarithmic equation \(\log _{4}\left(x^{2}+x\right)-\log _{4}(x+1)\) can be combined into one logarithm as \(\log_{4}\left(\frac{x^2+x}{x+1}\right)\).

Step 2: Convert logarithmic form to exponential form

The equation is now in the form that \(\log _{4}\left(\frac{x^2+x}{x+1}\right) = 2\) can be converted to exponential form according to the definition of a logarithm. As a result, \(\left(\frac{x^2+x}{x+1}\right) = 4^2\), which simplifies to \(\left(\frac{x^2+x}{x+1}\right) = 16\).

Step 3: Solving the equation for \(x\)

Now, the equation \(\left(\frac{x^2+x}{x+1}\right) = 16\) is a rational equation, which can be solved by multiplying each side by \((x+1)\), then simplifying, to find the quadratic equation \(x^2 + x = 16(x+1)\). Solve this equation to find the values of \(x\).

Step 4: Check the solution

After obtaining the solutions of quadratic equation, check each to see if they satisfy the original equation.

Key Concepts

Quotient Rule for Logarithms

When dealing with logarithmic equations, one useful tool is the quotient rule for logarithms. This rule helps simplify expressions where one logarithm is subtracted from another with the same base. In mathematical terms, the quotient rule is stated as:
\[\begin{equation}\log_b(A) - \log_b(B) = \log_b\left(\frac{A}{B}\right)\end{equation}\]
In practice, this means if you have \(\log_b{A} - \log_b{B}\), you can combine them into a single logarithm that's the log of the division of \(A\) by \(B\). This is particularly handy as it sets the stage for converting the logarithmic form to the exponential form, a necessary step in solving many logarithmic equations.

Converting Logarithmic Form to Exponential Form

One of the central skills when dealing with logarithms is knowing how to switch between logarithmic and exponential forms. This is a powerful step in the process of solving logarithmic equations. Here's the fundamental relationship between the two:
\[\begin{equation}\log_b(A) = C \Leftrightarrow b^C = A\end{equation}\]
This means if you have \(\log_b(A) = C\), it's equivalent to saying that \

Solving Rational Equations

A rational equation is one that involves at least one rational expression—an algebraic fraction whose numerator and/or denominator contain variables. In the context of solving such equations, the goal is to isolate the variable and solve for its value. Common steps in this process often include finding a common denominator to combine terms, cross-multiplying to eliminate fractions, and then simplifying the resulting equation. As rational equations might yield solutions that are not valid (because they can make the denominator equal to zero, which is undefined), it is crucial to check all potential solutions back in the original equation to ensure they are valid. This verification step is an essential part of solving rational equations, as it confirms the legitimacy of the solutions.