X2+y2=7xythen prove that log x+y/3=1/2logx+logy

1.	X2+y2=7xythen prove that log x+y/3=1/2logx+logy
Answer» ANSWER: QUESTION: x² + y² = 7xy prove that x+y/3=1/2logx+logy Solution: $: { \implies{ \sf{ {x}^{2} + {y}^{2} = 7xy }}} \\ \\ { \sf{Adding \: 2xy \: on \: both \: sides}} \\ \\ : { \implies{ \sf{ {x}^{2} + {y}^{2} + 2xy = 7xy + 2xy }}} \\ \\ : { \implies{ \sf{ {(x + y)}^{2} = 9xy }}} \\ \\ : { \implies{ \sf{(x + y) = \sqrt{9xy} }}} \\ \\ : { \implies{ \sf{(x + y) = 3 \sqrt{xy} }}} \\ \\ : { \implies{ \sf{ \frac{x + y}{3} = {(xy)}^{ \frac{1}{2} } }}} \\ \\ { \sf{Taking \: Log \: On \: Both \: Sides}} \\ \\ : { \implies{ \sf{log \bigg( \frac{x + y}{3} \bigg) = log {(xy)}^{ \frac{1}{2} } }}} \\ \\ : { \implies{ \sf{log \bigg( \frac{x + y}{3} \bigg) = \frac{1}{2} (logxy)}}} \\ \\ : { \implies{ \sf{log \bigg( \frac{x + y}{3} \bigg) = \frac{1}{2} (logx + logy)}}} \: \\ \\ \rm \: Hence \: Proved$ Used FORMULAE: $\boxed{ \sf{log {a}^{n} = n \: loga}}$ $\boxed{ \sf{logab = loga + logb}}$

Answer»

QUESTION:

x² + y² = 7xy prove that x+y/3=1/2logx+logy

Solution:

$: { \implies{ \sf{ {x}^{2} + {y}^{2} = 7xy }}} \\ \\ { \sf{Adding \: 2xy \: on \: both \: sides}} \\ \\ : { \implies{ \sf{ {x}^{2} + {y}^{2} + 2xy = 7xy + 2xy }}} \\ \\ : { \implies{ \sf{ {(x + y)}^{2} = 9xy }}} \\ \\ : { \implies{ \sf{(x + y) = \sqrt{9xy} }}} \\ \\ : { \implies{ \sf{(x + y) = 3 \sqrt{xy} }}} \\ \\ : { \implies{ \sf{ \frac{x + y}{3} = {(xy)}^{ \frac{1}{2} } }}} \\ \\ { \sf{Taking \: Log \: On \: Both \: Sides}} \\ \\ : { \implies{ \sf{log \bigg( \frac{x + y}{3} \bigg) = log {(xy)}^{ \frac{1}{2} } }}} \\ \\ : { \implies{ \sf{log \bigg( \frac{x + y}{3} \bigg) = \frac{1}{2} (logxy)}}} \\ \\ : { \implies{ \sf{log \bigg( \frac{x + y}{3} \bigg) = \frac{1}{2} (logx + logy)}}} \: \\ \\ \rm \: Hence \: Proved$

Used FORMULAE:

$\boxed{ \sf{log {a}^{n} = n \: loga}}$

$\boxed{ \sf{logab = loga + logb}}$