Solution set of the inequality is:

Explanation: 1. Define the Domain: For the logarithms to be defined: The common domain is . 2. Simplify the Inequality: Use the base change property and the product rule. Left Side (LHS): Combining with the first term: Using : Right Side (RHS): Using : 3. Solve the resulting inequality: Since the base ( ) is greater than 1, the inequality direction remains the same: 4. Intersection with Domain: We must satisfy both (from the domain) and . In interval notation, this is . The correct option is (b) .

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NIMCET 2011 — Mathematics PYQ

NIMCET | Mathematics | 2011

Solution set of the inequality $\log_{3}(x + 2)(x + 4) + \log_{1/3}(x + 2) < \frac{1}{2}\log_{\sqrt{3}}7$ is:

Choose the correct answer:

Explanation

1. Define the Domain:

For the logarithms to be defined:

$x + 2 > 0 \Rightarrow x > -2$
$x + 4 > 0 \Rightarrow x > -4$

The common domain is $x > -2$ .

2. Simplify the Inequality:

Use the base change property $\log_{a^n}b = \frac{1}{n}\log_{a}b$ and the product rule.

Left Side (LHS):

$\log_{1/3}(x + 2) = \log_{3^{-1}}(x + 2) = -\log_{3}(x + 2)$
Combining with the first term: $\log_{3}[(x + 2)(x + 4)] - \log_{3}(x + 2)$
Using $\log M - \log N = \log(\frac{M}{N})$ :

$\text{LHS} = \log_{3}\left(\frac{(x + 2)(x + 4)}{x + 2}\right) = \log_{3}(x + 4)$

Right Side (RHS):

$\frac{1}{2}\log_{\sqrt{3}}7 = \frac{1}{2}\log_{3^{1/2}}7$
Using $\log_{3^{1/2}}7 = \frac{1}{1/2}\log_{3}7 = 2\log_{3}7$ :

$\text{RHS} = \frac{1}{2}(2\log_{3}7) = \log_{3}7$

3. Solve the resulting inequality:

\log_{3}(x + 4) < \log_{3}7

Since the base ( $3$ ) is greater than 1, the inequality direction remains the same:

x + 4 < 7

x < 3

4. Intersection with Domain:

We must satisfy both $x > -2$ (from the domain) and $x < 3$ .

-2 < x < 3

In interval notation, this is $(-2, 3)$ .

The correct option is (b) $(-2, 3)$ .