Question

If x2+y2 = 14xy then p/T 2 long [x+y] =4 long 2 + log x + logy

Answer 1

✿$\underline{\textbf{\textsf{ \purple{Solution}:- }}}$

$\sf{\\}$

$\sf{ (x + y)^2 = x^2 + y^2 + 2xy }$

$\longrightarrow \sf{ (x + y)^2 = 14xy + 2xy }$

$\longrightarrow \sf{ (x + y)^2 = 16xy \quad \quad \dots(1)}$

$\sf{\\}$

$\underline{\texttt{ \red{To\:prove}:- }}$

$\sf{ 2log(x+y) = 4log2 + logx + logy}$

$\sf{\\}$

$\underline{\textbf{ \pink{L.H.S}:- }}$

$\sf{ 2 log (x + y)}$

${\large{\green{\bigstar}}} {\boxed{\sf{a log(b) = {log(b)}^{a}}}}$

$\longrightarrow \sf{ log (x + y)^2}$

$\longrightarrow \sf{ log (16xy) \quad \quad [From\:(1)]}$

${\large{\red{\bigstar}}} {\boxed{\sf{ log(ab) = log(a) + log(b)}}}$

$\longrightarrow \sf{ log16 + log(x) + log(y) }$

$\longrightarrow \sf{ log(2)^4 + log(x) + log(y)}$

${\large{\blue{\bigstar}}} {\boxed{\sf{ log(b)^a = a\:log(b)}}}$

$\longrightarrow \sf{ 4\:log(2) + log(x) + log(y) }$

$\sf{\\}$

$\underline{\textbf{ \pink{R.H.S}:- }}$

$\sf{ 4\:log(2) + log(x) + log(y) }$

$\sf{\\}$

$\leadsto \underline{\boxed{\mathfrak{\sf{ \pink{L.H.S\:=\:R.H.S} }}}}$

Hence proved.