\[\begin{array}{l}
Q4. \\
& \left( a \right)\,\,{3^{2{x^2}}} - 2 \cdot {3^{{x^2} + x + 6}} + {3^{2x + 12}} = 0 \\
& \left( b \right)\,\,\,{x^2} \cdot {2^{\sqrt {2x + 1} - 1}} + {2^x} = {2^{\sqrt {2x + 1} + 1}} + {x^2} \cdot {2^{x - 2}} \\
& \left( c \right)\,\,\,{\left( {5{x^2} - 6} \right)^{\frac{1}{4}}} = x \\
\end{array}\]
\[\begin{array}{l}
Q5. \\
& \left( a \right)\,\,{4^{{{\log }_{64}}\left( {x - 3} \right) + {{\log }_2}5}} = 50 \\
& \left( b \right)\,\,\,{x^{\log {{\left( {1 - x} \right)}^2}}} = 9 \\
& \left( c \right)\,\,\,{\log _3}x + {\log _9}x + {\log _{27}}x = 5.5 \\
& \left( d \right)\,\,\,{\log _2}\left( {3 - x} \right) + {\log _2}\left( {1 - x} \right) = 3 \\
& \left( e \right)\,\,\,\log \left( {x - 3} \right) + \log \left( {x + 6} \right) = \log 2 + \log 5 \\
& \left( f \right)\,\,\,\,\,\log \left( {x - 4} \right) + \log \left( {x + 3} \right) = \log \left( {5x + 4} \right) \\
& \left( g \right)\,\,\,\,\ln \left( {{x^2} + 1} \right) - \frac{1}{2}\ln \left( {{x^2} + 2x + 1} \right) = \ln 3 \\
& \left( h \right)\,\,\,{\log _5}\left( {x - 2} \right) + 2{\log _5}\left( {{x^2} - 2} \right) + {\log _5}{\left( {x - 2} \right)^{ - 1}} = 4 \\
& \left( i \right)\,\,\,{\log _2}\frac{{x - 2}}{{x - 1}} - 1 = {\log _2}\frac{{3x - 7}}{{3x - 1}} \\
& \left( j \right)\,\,2\,{\log _2}\frac{{x - 7}}{{x - 1}} + {\log _2}\frac{{x - 1}}{{x + 1}} = 1 \\
& \left( k \right)\,\,{\log _3}\left( {5x - 2} \right) - 2{\log _3}\sqrt {3x + 1} = 1 - {\log _3}4 \\
& \left( l \right)\,\,\log \left( {3x - 2} \right) - 2 = \frac{1}{2}\log \left( {x + 2} \right) - \log 50 \\
& \left( m \right)\,\,{\log ^2}\left( {1 + \frac{4}{x}} \right) + {\log ^2}\left( {1 - \frac{4}{{x + 4}}} \right) = 2{\log ^2}\left( {\frac{2}{{x - 1}} - 1} \right) \\
\end{array}\]
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