Math, asked by kritikriti055, 5 months ago

If log(x+y/2) = 1/2(log x+log y) show that x=y​

Answers

Answered by Arceus02
1

Given:-

  •   \sf \: log \bigg( \dfrac{x + y}{2}  \bigg) =  \dfrac{1}{2}  \bigg( log(x)  +  log(y)  \bigg)

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To show:-

  • x = y

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Answer:-

Given that,

  \sf \: log \bigg( \dfrac{x + y}{2}  \bigg) =  \dfrac{1}{2}  \bigg( log(x)  +  log(y)  \bigg)

{\red{\bigstar}} \boxed{\sf{log\bigg(\dfrac{a}{b}  \bigg) = log(a) - log(b)}}

   \sf \longrightarrow \: log(x + y) - log(2) =  \dfrac{1}{2}  \bigg( log(x)  +  log(y)  \bigg)

{\blue{\bigstar}} \boxed{\sf{log(a) + log(b) = log(ab)}}

   \sf \longrightarrow \: log(x + y) - log(2) =  \dfrac{1}{2}  \times  log(xy)

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

   \sf \longrightarrow \: log(x + y) - log(2) =   log(xy {}^{1 / 2} )

   \sf \longrightarrow \: log(x + y) -     log( \sqrt{xy}  )    =  log(2)

{\red{\bigstar}} \boxed{\sf{log\bigg(\dfrac{a}{b}\bigg) = log(a) - log(b)}}

   \sf \longrightarrow \: log \bigg( \dfrac{x + y}{ \sqrt{xy} }  \bigg)    =  log(2)

Comparing both sides,

 \sf \longrightarrow   \dfrac{x + y}{ \sqrt{xy} }  = 2

 \sf \longrightarrow   x + y  = 2 \sqrt{xy}

Squaring both sides,

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

 \sf \longrightarrow  {x}^{2}  +  {y}^{2}  + 2xy = 4xy

 \sf \longrightarrow  {x}^{2}  +  {y}^{2}   -  2xy = 0

 \sf \longrightarrow  {(x - y)}^{2}  = 0

 \sf \longrightarrow  (x - y)(x - y) = 0

So,

 \longrightarrow  \underline{ \underline{ \sf{ \green{   x = y  }}}}

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