Correct Answer - Option 3 : 1/2
Concept:
Formula of Logarithms:
\({a^b} = x\; \Leftrightarrow lo{g_a}x = b\), here a ≠ 1 and a > 0 and x be any number.
Properties of Logarithms:
- \({\log _a}a = 1\)
- \({\log _a}\left( {x.y} \right) = {\log _a}x + {\log _a}y\)
- \({\log _a}\left( {\frac{x}{y}} \right) = {\log _a}x - {\log _a}y\)
- \({\log _a}\left( {\frac{1}{x}} \right) = - {\log _a}x\)
- \({\rm{lo}}{{\rm{g}}_a}{x^p} = p{\rm{lo}}{{\rm{g}}_a}x\)
- \(lo{g_a}\left( x \right) = \frac{{lo{g_b}\left( x \right)}}{{lo{g_b}\left( a \right)}}\)
Calculation:
Given: \(3{\log _4}2 - \frac{1}{2}{\log _3}9\)
\(\Rightarrow 3{\log _4}2 - \frac{1}{2}{\log _3}9 = 3{\log _{{2^2}}}2 - \frac{1}{2}{\log _3}{3^2}\)
Using the rule, \(lo{g_{{a^b}}}\left( x \right) = \frac{1}{b}lo{g_a}\left( x \right)\) in \({\log _{{2^2}}}2\) we get,
\(\Rightarrow {\log _{{2^2}}}2 = \frac{1}{2}{\log _2}2\)
Also, by using the power rule in \({\log _3}{3^2}\) we get,
\(\Rightarrow {\log _3}{3^2} = 2{\log _3}3\)
\(\Rightarrow 3{\log _4}2 - \frac{1}{2}{\log _3}9 = \frac{3}{2}{\log _2}2 - \frac{2}{2}{\log _3}3\)
\(= \frac{3}{2}{\log _2}2 - {\log _3}3\)
As we know that, \({\log _3}3 = {\log _2}2 = 1\)
= (3/2) - 1
= 1/2
Hence,
\(3{\log _4}2 - \frac{1}{2}{\log _3}9 = \frac{1}{2}\).