NIMCET 2019 — Mathematics PYQ

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Question

NIMCET 2019 — Mathematics PYQ

NIMCET | Mathematics | 2019

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

Choose the correct answer:

Explanation

Concept:
- $lo g_{α^{β}} m = \frac{1}{β} lo g_{α} m$
- $lo g_{α} \frac{m}{n} = lo g_{α} m - lo g_{α} n$
- If $lo g_{α} m = lo g_{α} n$ then $m = n$

Calculation:
$\log_{3}(x+2)(x+4)+\log_{\frac{1}{3}}(x+2)<\frac{1}{2}\log_{\sqrt{3}}7$

As we know that, $lo g_{α} m = \frac{1}{β} lo g_{α} m$
$\log_{3} (x + 2) (x + 4) - \log_{3} (x + 2) < \log_{3} 7$
As we know that, $lo g_{α} \frac{m}{n} = lo g_{α} m - lo g_{α} n$
$\log_{3} (x + 4) < \log_{3} 7$
As we know that, if $lo g_{α} m = lo g_{α} n$ then $m = n$
$\Rightarrow x + 4 < 7$
$\Rightarrow x < 3$ ------------(1)
For $lo g_{3} (x + 2) (x + 4)$ to exist $x \in (- 2, \infty)$ -------------(2)
For $lo g_{3} (x + 2) \Rightarrow x \in (- 2, \infty)$ -------------(3)
By combining equation 1, 2, 3 we get $x \in (- 2, 3)$