Math, asked by Anonymous, 1 day ago

By using LHS, prove that it is equal to RHS.

 \dfrac12 \ln(\pi) + \dfrac12 \ln(2) = \ln(\sqrt{2\pi})

Answers

Answered by piyush09s
4

LHS

 =  \frac12 \ln(\pi) + \frac12 \ln(2)  \\   \\ =  \frac{1}{2} \left[ \ln(\pi) +  \ln(2) \right] \\  \\  =  \frac{1}{2} \left[  \ln(2\pi)\right] \\  \\ =  \left[  \ln(2\pi)^{ \frac{1}{2} } \right] \\  \\  =  \ln \sqrt{2\pi}

I'VE USED PROPERTIES WHICH ARE

ln(x) + ln(y) = ln(xy)

yln(x) = ln(x^y)

Please thank my answer and mark brainliest.

Answered by mathdude500
8

\large\underline{\sf{Solution-}}

Consider LHS

\rm :\longmapsto\:\dfrac{1}{2}ln(\pi )+  \dfrac{1}{2}ln(2)

\rm \:  =  \: \dfrac{1}{2}[ln(\pi) + ln(2)]

We know,

\boxed{\tt{ ln(x) + ln(y) = ln(xy)}}

So, using this identity, we get

\rm \:  =  \: \dfrac{1}{2}ln(2\pi)

We know,

\boxed{\tt{ yln(x) = ln( {x}^{y})}}

So, using this identity, we get

\rm \:  =  \: ln \sqrt{2\pi}

Hence,

 \purple{\rm :\longmapsto\:\boxed{\tt{ \dfrac{1}{2}ln(\pi )+  \dfrac{1}{2}ln(2) = ln( \sqrt{2\pi})}}}

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Note :-

Proof of Logarithmic Property Used

 \purple{\rm :\longmapsto\:(1). \:  log_{a}( {x}^{n} ) = n log_{a}(x)}

Let assume that

\rm :\longmapsto\: log_{a}(x) = y

\rm :\longmapsto\:x =  {a}^{y}

\rm\implies \: {x}^{n} =  {a}^{yn}

\rm\implies \: log_{a}{x}^{n} =  ny

\bf\implies \: log_{a}{x}^{n} =  n log_{a}(x)

Hence, Proved

 \purple{\rm :\longmapsto\:(2). \:  log_{a}( {x})  +  log_{a}(y) =  log_{a}(xy)}

Let assume that

\rm :\longmapsto\: log_{a}(x) = m\rm\implies \:x =  {a}^{m}  -  -  - (1)

and

\rm :\longmapsto\: log_{a}(y) = n\rm\implies \:y =  {a}^{n}  -  -  - (2)

So,

\rm :\longmapsto\:xy =  {a}^{m} \times  {a}^{n}

\rm :\longmapsto\:xy =  {a}^{m + n}

\rm\implies \: log_{a}(xy) = m + n

\bf\implies \: log_{a}(xy) =  log_{a}(x)  +  log_{a}(y)

Hence, Proved

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