Correct Answer - Option 2 : (-2, 3)
Concept:
- \({\log _{{\alpha ^\beta }}}m = \frac{1}{\beta }{\log _\alpha }m\)
- \({\log _\alpha }\frac{m}{n} = {\log _\alpha }m - {\log _\alpha }n\)
- \(If \ {\log _\alpha }m = {\log _\alpha }n\;\ then\;m = n\)
Calculation:
\(\log_3 (x + 2)(x + 4) + \log_{\frac 1 3} (x + 2) < \frac 1 2 \log_\sqrt 3 7\)
As we know that, \({\log _{{\alpha ^\beta }}}m = \frac{1}{\beta }{\log _\alpha }m\)
\({\log _3}\left( {x + 2} \right)\left( {x + 4} \right) - {\log _3}\left( {x + 2} \right) < {\log _3}7\)
As we know that, \({\log _\alpha }\frac{m}{n} = {\log _\alpha }m - {\log _\alpha }n\)
\({\log _3}\left( {x + 4} \right) < {\log _3}7\)
As we know that, \(if \ {\log _\alpha }m = {\log _\alpha }n\;\ then\;m = n\)
⇒x + 4 < 7
⇒ x < 3 ------------(1)
For \({\log _3}\left( {x + 2} \right)\left( {x + 4} \right)\) to exist \(x ∈ ( - 2, \infty )\)---------(2)
For \({\log _3}\left( {x + 2} \right) ⇒ x ∈ ( - 2, \infty )\)-----------------(3)
By combining equation 1, 2, 3 we get \(x ∈ ( - 2,3)\)
Hence, option B is the correct answer.
