Question

if (x-y)^2=121xy show that 2 log(x+y)=3log5+logx+logy...Please answer this question.... Fast​

Answer 1

Step-by-step explanation:

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Answer 2

$\textsf{\large{\underline{Solution}:}}$

Given That:

$\rm \longrightarrow {(x - y)}^{2} = 121xy$

$\rm \longrightarrow {x}^{2} + {y}^{2} - 2xy = 121xy$

Adding 4xy to both sides, we get:

$\rm \longrightarrow {x}^{2} + {y}^{2} - 2xy + 4xy = 121xy + 4xy$

$\rm \longrightarrow {x}^{2} + {y}^{2} + 2xy = 125xy$

$\rm \longrightarrow {(x + y)}^{2}= 125xy$

Taking log on both sides, we get:

$\rm \longrightarrow \log{(x + y)}^{2}= \log(125xy)$

$\rm \longrightarrow 2\log(x + y)= \log(125) + \log(x) + \log(y)$

$\rm \longrightarrow 2\log(x + y)= \log( {5}^{3} ) + \log(x) + \log(y)$

$\rm \longrightarrow 2\log(x + y)= 3\log( {5}^{} ) + \log(x) + \log(y)$

Hence Proved..!!

$\textsf{\large{\underline{Learn More}:}}$

$\rm 1. \: \: {a}^{n} = b \implies log_{a}(b) = n$

$\rm 2. \: \: log_{a}(1) = 0, \: a \neq0,1$

$\rm 3. \: \: log_{a}(a) = 1, \: a \neq0,1$

$\rm 4. \: \: log_{a}(x) = log_{a}(y) \implies x = y$

$\rm 5. \: \: log_{e}(x) = ln(x)$

$\rm6. \: \: log_{a}(x) + log_{a}(y) = log_{a}(xy)$

$\rm7. \: \: log_{a}(x) - log_{a}(y) = log_{a} \bigg( \dfrac{x}{y} \bigg)$

$\rm 8. \: \: log_{a}( {x}^{n} ) = n\log_{a}(x)$

$\rm 9. \: \: log_{a}(m) = \dfrac{ log_{b}(m) }{ log_{b}(a) },m > 0,b > 0,a \ne1,b \ne1$

$\rm 10. \: \: log_{a}(b) = \dfrac{1}{ log_{b}(a) }$