Question

1/2 log(x+y) -1/2 log(x-y) as single logarithms​

Answer 1

$\rm \longrightarrow\dfrac{1}{2} log(x+y) -\dfrac{1}{2} \: log(x-y) \\ \\ \\ \rm \longrightarrow log( {x+y} ) ^{ \frac{1}{2} } - \:log( {x+y} ) ^{ \frac{1}{2}}\\ \\ \\ \rm \longrightarrow log \frac{( {x+y} ) ^{ \frac{1}{2} }}{( {x+y} ) ^{ \dfrac{1}{2}}} \\ \\ \\ \rm \longrightarrow log \sqrt{ \frac{( {x+y} ) ^{ }}{( {x - y} ) ^{ }} }$

Learn more :

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