Question

Log 3 base 4 * log 4base 3 + log 32 base 2

Answer 1

Given Expression :-

$\\ \quad \bullet \quad \sf log_{4}(3) \times log_{3}(4) + log_{2}(32)$

Properties To Be Used :-

$\\ \quad \mapsto \quad \sf \dfrac{1}{ log_{b}(a) } = log_{a}(b)$

$\\ \quad \mapsto \quad \sf log_{a}( {b}^{c} ) = c \times log_{a}(b)$

$\\ \quad \mapsto \quad \sf log_{a}(a) = 1$

Calculation :-

$\\ \quad \longrightarrow \quad \sf log_{4}(3) \times log_{3}(4) + log_{2}(32)$

$\\ \quad \longrightarrow \quad \sf \dfrac{1}{ \cancel{ log_{3}(4)} } \times \cancel{log_{3}(4)} + log_{2}( {2}^{5} )$

$\\ \quad \longrightarrow \quad \sf 1 + 5 \times log_{2}(2)$

$\\ \quad \longrightarrow \quad \sf 1 + 5(1)$

$\\ \quad \longrightarrow \quad \sf 1 + 5$

$\\ \quad \therefore \quad \boxed{ \frak{6}} \bigstar$

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Answer 2

Answer:

$\footnotesize \implies log_{4}(3) \times log_{4}(3) + log_{2}(32)$

Important Properties of Log:

$\footnotesize \implies log_{x}(y) = \dfrac{1}{ log_{y}(x) } \: or \: \dfrac{ log_{e}(y) }{ log_{e}(x) }$

$\footnotesize \implies log_{x}(y) \times log_{y}(x) = 1$

$\footnotesize \implies log_{e}( \frac{m}{n} ) = log_{e}(m) - log_{e}(n)$

$\footnotesize \implies log_{e}(mn) = log_{e}(m) + log_{e}(n)$

$\footnotesize \implies log(m^n) = n\:log\:m$

$\footnotesize \implies log_{a}(a) = 1$

Required Solution:

$\footnotesize \implies log_{4}(3) \times log_{3}(4) + log_{2}(32)$

$\footnotesize \bullet \: \bf log_{3}(4) = \dfrac{1}{ log_{4}(3) }$

$\footnotesize \implies log_{4}(3) \times \dfrac{1}{ log_{4}(3) } + log_{2}(32)$

$\footnotesize \implies 1 + log_{2}(32)$

$\footnotesize \bullet \: \bf log_{2}(32) = x$

$\footnotesize \bullet \: \bf {2}^{x} = 32$

Let x = 5

$\footnotesize \bullet \: \bf {2}^{5} = 32$

Therefore,

$\footnotesize \bullet \: \bf log_{2}(32) = 5$

$\footnotesize \implies 1 + 5$

$\footnotesize \implies \underline{ \boxed{ \red{ 6}}}$