Question

please tell me fast and please give correct answer solve it with the proper full calculation .​

Answer 1

Given Question :-

If

$\rm :\longmapsto\:log\bigg[\dfrac{a - b}{2} \bigg] \: = \dfrac{1}{2}\bigg[loga + logb\bigg]$

then, prove that

$\rm :\longmapsto\: {a}^{2} + {b}^{2} = 6ab$

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

Given that,

$\rm :\longmapsto\:log\bigg[\dfrac{a - b}{2} \bigg] \: = \dfrac{1}{2}\bigg[loga + logb\bigg]$

can be rewritten as

$\rm :\longmapsto\:2log\bigg[\dfrac{a - b}{2} \bigg] \: = \bigg[loga + logb\bigg]$

We know,

$\boxed{ \tt{ \: bloga = log {a}^{b} \: \: }}$

and

$\boxed{ \tt{ \: loga + logb = log(ab) \: \: }}$

So, using this, we get

$\rm :\longmapsto\:log {\bigg[\dfrac{a - b}{2} \bigg]}^{2} = log(ab)$

So, on comparing, we get

$\rm :\longmapsto\: {\bigg[\dfrac{a - b}{2} \bigg]}^{2} = ab$

$\rm :\longmapsto\:\dfrac{ {a}^{2} + {b}^{2} - 2ab}{4} = ab$

$\rm :\longmapsto\: {a}^{2} + {b}^{2} - 2ab = 4ab$

$\rm :\longmapsto\: {a}^{2} + {b}^{2} = 2ab + 4ab$

$\red{\sf :\longmapsto\: {a}^{2} + {b}^{2} = 6ab \: \: }$

Hence, Proved

Additional Information :-

$\boxed{ \tt{ \: log_{x}(x) = 1 \: \: }}$

$\boxed{ \tt{ \: log_{ {x}^{a} }( {x}^{b} ) = \frac{b}{a} \: \: }}$

$\boxed{ \tt{ \: {e}^{logx} = x \: \: }}$

$\boxed{ \tt{ \: {e}^{ylogx} = {x}^{y} \: \: }}$

$\boxed{ \tt{ \: {a}^{ log_{a}(x) } = x \: \: }}$

$\boxed{ \tt{ \: {a}^{ ylog_{a}(x) } = {x}^{y} \: \: }}$