Question

$\spadesuit\ \; \; \; \bf \orange{Maths}\ \; :$
Real numbers x , y satisfy
$\rm \green{2^x = 3^y = 216\ .\ Find\ \dfrac{1}{x}+\dfrac{1}{y}\ .}$
$\bf . \qquad Best\ of\ Luck\ ! \qquad .$​

Answer 1

Answer:

Given :-

$\tt {2}^{x} = {3}^{y} = 216 \: \: \: ( x , y ) \in \mathbb R2$

To Find :-

$\sf \: The \: value \: of \: \tt \dfrac{1}{x} + \dfrac{1}{y}$

Solution :-

By Given we have ;

$\tt {2}^{x} = 216 \quad and \quad {3}^{y} = 2162$

Now , Consider ;

$\quad \qquad { \tt { {2}^{x} = 216 } }$

Apply Prime Factorisation on RHS we have ;

$\quad \qquad { : \longmapsto \tt { {2}^{x} = 2 × 2 × 2 × 3 × 3 × 3} }:$

It can be written further as ;

$\quad \qquad { : \longmapsto \tt { 2^{x} = 6 × 6 × 6 }}:$

$\quad \qquad { : \longmapsto \tt { 2^{x} = 6³ }}:$

Now Take log on both sides ;

$\quad \qquad { : \longmapsto \tt { log ( 2^{x} ) = log ( 6³ ) }}:$

We knows a Logarithmic identity i.e ;

$\quad \qquad { \bigstar { \underline { \boxed { \bf { \red { log ( m^{n} ) = n log ( m ) }}}}}}{\bigstar}$

Using this we have ;

$\quad \qquad { : \longmapsto \tt { x . log ( 2 ) = 3 . log ( 6 ) }}:$

Can be written as ;

$\quad \qquad { : \longmapsto \tt { x = \dfrac{3 . log( 6 )}{log ( 2 )} }}:$

Again Can be written as ;

$\quad \qquad { : \longmapsto \tt { \dfrac{x}{1} = \dfrac{3 . log( 6 )}{log ( 2 )} }}:$

Reciprocal both sides we have ;

$\quad \qquad { : \longmapsto \tt {\green { \dfrac{1}{x} = \dfrac{log ( 2 )}{ 3 . log ( 6 )} \quad ----( i ) }}}:$

Now , Consider ;

$\quad \qquad { \tt { {3}^{y} = 216 } }$

Again applying prime Factorisation as above we have ;

$\quad \qquad { : \longmapsto \tt { 3^{y} = 6³ }}:$

Taking log on both sides we have ;

$\quad \qquad { : \longmapsto \tt { log ( 3^{y} ) = log ( 6³ ) }}:$

Using the above identity as used above we have ;

$\quad \qquad { : \longmapsto \tt { y . log ( 3 ) = 3 . log ( 6 ) } }:$

$\quad \qquad { : \longmapsto \tt { y = \dfrac{3 . log ( 6 )}{log ( 3 ) } }}:$

Can be written as ;

$\quad \qquad { : \longmapsto \tt { \dfrac{y}{1} = \dfrac{3 . log ( 6 )}{log ( 3 ) } }}:$

Reciprocal both sides we have ;

$\quad \qquad { : \longmapsto \tt { \green { \dfrac{1}{y} = \dfrac{ log ( 3 )}{ 3 . log ( 6 ) } \quad ----( ii ) }}}:$

Adding ( i ) & ( ii ) we have ;

$\quad \qquad { : \longmapsto \tt { \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{ log ( 2 )}{ 3 . log ( 6 )} + \dfrac{ log ( 3 ) }{3 . log ( 6 )} }}:$

As the Denominators are same of both fractions we can simply add their numerators with same base ;

$\quad \qquad { : \longmapsto \tt { \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{ log ( 2 ) + log ( 3 )}{ 3 . log ( 6 )} }}:$

We knows another Logarithmic identity i.e ;

$\quad \qquad { \bigstar { \underline { \boxed { \bf { \red { log ( a ) + log ( b ) = log ( a.b) }}}}}}{\bigstar}$

Using this we have ;

$\quad \qquad { : \longmapsto \tt { \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{ log ( 2 . 3)}{ 3 . log ( 6 )} }}:$

$\quad \qquad { : \longmapsto \tt { \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{ log ( 6 )}{ 3 . log ( 6 )} }}:$

After Cancelling log ( 6 ) from both Numerator & Denominator we have ;

$\quad \qquad { : \longmapsto \tt { \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{3} }}:$

$\quad \qquad { \bigstar { \underline { \boxed { \bf { \red { \therefore { \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{3} }}}}}}}{\bigstar}$

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