Question

$log_{2}(32) - log_{4}(32)$
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Answer 1

$\tt \longrightarrow log_{2}(32) - log_{4}(32) \\ \\ \tt \longrightarrow log_{2}( {2}^{5} ) - log_{ {2}^{2} }( {2}^{5} )\\ \\ \tt \longrightarrow5 \: log_{2}( {2} ) - \dfrac{5}{2} log_{ {2} }( {2} )\\ \\ \tt \longrightarrow(5 \times 1)-( \dfrac{5}{2} \times 1)\\ \\ \tt \longrightarrow5 - \dfrac{5}{2}\\ \\ \tt \longrightarrow \dfrac{10 - 5}{2} = \dfrac{ 5}{2}$

Answer : 5/2

$\bf Additional- Information :$

$\boxed{\boxed{\begin{minipage}{4.2cm}\circ\sf\:\ln 1=0\\\circ\sf\:\ln e=1\\\circ\sf\:\ln x=y\leftrightarrow e^y=x\\\circ\sf\:e^{\ln x}=x,\:x>0\\\circ\sf\:\ln(e^x)=x,\:x\in\mathbb{\large R}\\\circ\sf\:\ln(xy)=\ln x+\ln y\\\circ\sf\:\ln(x/y)=\ln x-\ln y\\\circ\sf\:\ln(x^r)=r\ln x\\\circ\sf\:\ln x=log_e\:x\end{minipage}}}$

$\boxed{\boxed{\begin{minipage}{5cm}\displaystyle\circ\sf\ ^{a} log \ a= 1\\\\\circ \ ^{a}log \ 1 = 0 \\\\\circ \ ^{a ^{n}} log \ b^{m}= \dfrac{m}{n} \times\:^{a}log \ b \\\\\circ \ ^{a^{m}} \ log \ b^{m} = \ ^{a}log \ b \\\\\circ \ ^{a}log \ b = \dfrac{1}{^{b}log \ a} \\\\\circ \ ^{a}log \ b = \dfrac{^{m}log \ b}{^{m} log \ a} \\\\\circ \ a^{^{a} logb} = b \\\\\circ \ ^{a}log \ b + ^{a}log \ c = \ ^{a}log(bc) \\\\\circ \ ^{a}log \ b -\: ^{a}log \ c = \ ^{a}log \left( \dfrac{b}{c} \right) \\\circ \ ^{a}log \ b \:\cdot\: ^{a}log \ c = \ ^{a}log \ c \\\\\circ \ ^{a}log \left( \dfrac{b}{c} \right) = \ ^{a}log \left(\dfrac{c}{b}\right)\end{minipage}}}$