Problem 309 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \log _{3}(7-x) \leq \frac{9}{16} \log _{2 \sqrt{2}}^{2} \frac{1}{4}+\log _{7-x} 9 $$

Step-by-Step Solution

Answer

The solution to the inequality is \(x = -\frac{73}{64}\)

Step 1 - Separating the inequality

The first step is to separate the given inequality into its two sides: \(\log_{3}(7-x)\) and \(\frac{9}{16} \log_{2 \sqrt{2}}^{2}\frac{1}{4}+\log _{7-x} 9\) respectively.

Step 2 - Simplifying right hand side

Simplify the right-hand side by first dealing with the expression \(\log_{2 \sqrt{2}}^{2}\) which simplifies to 1 (since \(2\sqrt{2}\) to the power of 2 is 8, and \(\log_8 8\) is 1), and then apply the multiplication property of logs: \[\frac{9}{16} \cdot 1 \cdot \frac{1}{4}+\log _{7-x} 9 = \frac{9}{64} + \log _{7-x} 9\]

Step 3 - Equating the bases

Now that the bases of the logarithmic elements on both sides are the same, write them as equal instead of as an inequality because it simplifies the calculation: \(7-x = \frac{9}{64} + 9\)

Step 4 - Solving for \(x\)

Solving the equation obtained in the third step for \(x\) and we get \(x = 7 - \frac{585}{64} = -\frac{73}{64}\)

Step 5 - Checking the solution

Substitute \(x = -\frac{73}{64}\) back into our original inequality to check whether the solution is correct or not.

Key Concepts

Logarithmic functions

Logarithmic functions are a way to express numbers based on levels of magnitude. When you see an equation with a log, it usually involves some exponential relationships. Here, the base of a log is crucial as it tells you what number you need to raise to a certain power to get another number. For example,

• If you have \(\log_{3}(7-x)\), it implies that raising 3 to some power should give you \(7-x\).
• Similarly, \(\log_{2\sqrt{2}}^2\frac{1}{4}\) highlights dealing with a base of \(2\sqrt{2}\), which is simplified to a base of 8, because \((2\sqrt{2})^2 = 8\).

The properties of logs come in handy, such as the multiplication property: \(\log_b(A \cdot B) = \log_b A + \log_b B\). This property helps in breaking down complex expressions into manageable parts.

Solving inequalities

When you solve inequalities involving logarithms, the goal is to find all values of \(x\) that satisfy the inequality condition. These are not equations where you find one specific value.

• For a log inequality like \(\log_{3}(7-x) \leq \frac{9}{16} \log_{2\sqrt{2}}^2 \frac{1}{4}\), the approach is to manipulate it algebraically to find the range of \(x\) values.
• Handling log expressions on both sides requires keeping track of their base. Remember, you usually only equate or solve log expressions when the bases are equal.

Always check the domain of your logarithmic functions because \(\log_b(a)\) is only defined if \(a > 0\). Hence, \(7-x\) must be greater than zero.

Simplifying expressions

Simplification is a core skill in making complex expressions more manageable and solving inequalities.

• Consider the expression \(\frac{9}{16} \log_{2\sqrt{2}}^2 \frac{1}{4}\). By realizing \((2\sqrt{2})^2 = 8\), the expression becomes simpler since \(\log_{8}(8) = 1\).
• The multiplication property helps break complicated log expressions. It states, \(a\log_b(c) = \log_b(c^a)\).

Simplifying the right side gives \(\frac{9}{64} + \log_{7-x} 9\), preparing the expressions for equating their bases.

Equating bases

Equating the bases is crucial when both sides of the inequality involve logs. If the logarithms have the same base, you can simplify your solving process by equating the arguments inside the logs directly.

• For the inequality \(\log_{3}(7-x) = \frac{9}{64} + \log_{7-x} 9\), adjusting for equal bases allows us to say directly \(7-x = \frac{9}{64} + 9\).
• Once you have equated the bases, solve the resulting equation for \(x\).

This simplification works because when you eliminate the logarithm, you are finding the exponent that satisfies the equation perfectly. Always remember to validate the solution in the original inequality to ensure it lies within the permissible domain of the function.