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Find the value of the expression \(3{\log _4}2 - \frac{1}{2}{\log _3}9\)
1. 3/2
2. 2/3
3. 1/2
4. 1/3
5. None of these

1 Answer

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Best answer
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:

  1. \({\log _a}a = 1\)
  2. \({\log _a}\left( {x.y} \right) = {\log _a}x + {\log _a}y\)
  3. \({\log _a}\left( {\frac{x}{y}} \right) = {\log _a}x - {\log _a}y\)
  4. \({\log _a}\left( {\frac{1}{x}} \right) = - {\log _a}x\)
  5. \({\rm{lo}}{{\rm{g}}_a}{x^p} = p{\rm{lo}}{{\rm{g}}_a}x\)
  6. \(lo{g_a}\left( x \right) = \frac{{lo{g_b}\left( x \right)}}{{lo{g_b}\left( a \right)}}\)

 

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, \(3{\log _4}2 - \frac{1}{2}{\log _3}9 = \frac{1}{2}\).

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