$\begin{aligned} \frac{1}{2} (\log x + \log y) will equal to \\ \log (\frac{x + y}{2}) if - \end{aligned}$ $\eqalign{ & \frac{1}{2}\left( {\log x + \log y} \right)\,\,{\text{will}}\,{\text{equal}}\,{\text{to}} \cr & \log \left( {\frac{{x + y}}{2}} \right)\,\,{\text{if - }} \cr}$

1y = 0
2x = √y
3x = y
4x = ^y/₂

Answer:- 1

Explanation:-

Solution:

\begin{aligned} \frac{1}{2} (\log x + \log y) = \log (\frac{x + y}{2}) \\ \Rightarrow \frac{1}{2} \log (x y) = \log (\frac{x + y}{2}) \\ \Rightarrow \log {(x y)}^{\frac{1}{2}} = \log (\frac{x + y}{2}) \\ \Rightarrow {(x y)}^{\frac{1}{2}} = (\frac{x + y}{2}) \\ \Rightarrow x y = {(\frac{x + y}{2})}^{2} \\ \Rightarrow 4 x y = x^{2} + y^{2} + 2 x y \\ \Rightarrow x^{2} + y^{2} - 2 x y = 0 \\ \Rightarrow {(x - y)}^{2} = 0 \\ \Rightarrow x - y = 0 \\ \Rightarrow x = y \end{aligned}

$\eqalign{ & \frac{1}{2}\left( {\log x + \log y} \right) = \log \left( {\frac{{x + y}}{2}} \right) \cr & \Rightarrow \frac{1}{2}\log \left( {xy} \right) = \log \left( {\frac{{x + y}}{2}} \right) \cr & \Rightarrow \log {\left( {xy} \right)^{\frac{1}{2}}} = \log \left( {\frac{{x + y}}{2}} \right) \cr & \Rightarrow {\left( {xy} \right)^{\frac{1}{2}}} = \left( {\frac{{x + y}}{2}} \right) \cr & \Rightarrow xy = {\left( {\frac{{x + y}}{2}} \right)^2} \cr & \Rightarrow 4xy = {x^2} + {y^2} + 2xy \cr & \Rightarrow {x^2} + {y^2} - 2xy = 0 \cr & \Rightarrow {\left( {x - y} \right)^2} = 0 \cr & \Rightarrow x - y = 0 \cr & \Rightarrow x = y\, \cr}$

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\begin{aligned} \frac{1}{2} (\log x + \log y) = \log (\frac{x + y}{2}) \\ \Rightarrow \frac{1}{2} \log (x y) = \log (\frac{x + y}{2}) \\ \Rightarrow \log {(x y)}^{\frac{1}{2}} = \log (\frac{x + y}{2}) \\ \Rightarrow {(x y)}^{\frac{1}{2}} = (\frac{x + y}{2}) \\ \Rightarrow x y = {(\frac{x + y}{2})}^{2} \\ \Rightarrow 4 x y = x^{2} + y^{2} + 2 x y \\ \Rightarrow x^{2} + y^{2} - 2 x y = 0 \\ \Rightarrow {(x - y)}^{2} = 0 \\ \Rightarrow x - y = 0 \\ \Rightarrow x = y \end{aligned}

\begin{aligned} \frac{1}{2} (\log x + \log y) = \log (\frac{x + y}{2}) \\ \Rightarrow \frac{1}{2} \log (x y) = \log (\frac{x + y}{2}) \\ \Rightarrow \log {(x y)}^{\frac{1}{2}} = \log (\frac{x + y}{2}) \\ \Rightarrow {(x y)}^{\frac{1}{2}} = (\frac{x + y}{2}) \\ \Rightarrow x y = {(\frac{x + y}{2})}^{2} \\ \Rightarrow 4 x y = x^{2} + y^{2} + 2 x y \\ \Rightarrow x^{2} + y^{2} - 2 x y = 0 \\ \Rightarrow {(x - y)}^{2} = 0 \\ \Rightarrow x - y = 0 \\ \Rightarrow x = y \end{aligned}

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$\begin{aligned} \frac{1}{2} (\log x + \log y) will equal to \\ \log (\frac{x + y}{2}) if - \end{aligned}$ $\eqalign{ & \frac{1}{2}\left( {\log x + \log y} \right)\,\,{\text{will}}\,{\text{equal}}\,{\text{to}} \cr & \log \left( {\frac{{x + y}}{2}} \right)\,\,{\text{if - }} \cr}$

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12(logx+logy)willequaltolog(x+y2)if - 12(log⁡x+log⁡y)willequaltolog⁡(x+y2)if - \eqalign{ & \frac{1}{2}\left( {\log x + \log y} \right)\,\,{\text{will}}\,{\text{equal}}\,{\text{to}} \cr & \log \left( {\frac{{x + y}}{2}} \right)\,\,{\text{if - }} \cr}

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NCERT 6 to 12

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$\begin{aligned} \frac{1}{2} (\log x + \log y) will equal to \\ \log (\frac{x + y}{2}) if - \end{aligned}$ $\eqalign{ & \frac{1}{2}\left( {\log x + \log y} \right)\,\,{\text{will}}\,{\text{equal}}\,{\text{to}} \cr & \log \left( {\frac{{x + y}}{2}} \right)\,\,{\text{if - }} \cr}$

NCERT
6 to 12