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

★ ════════════════════ ★

$\sf 2^x=216\ \qquad 3^y=216$

Applying logarithm on both sides ,

$\\ \longrightarrow \sf \log (2^x) = \log (216)\ \quad \log (3^y)= \log (216)$

$\longrightarrow \sf \log (2^x) = \log (6^3)\ \quad \log (3^y)= \log (6^3)$

$\boxed{ \bullet\ \; \sf \red{\log a^b = b\ \log a}}$

$\longrightarrow \sf x \log 2 = 3 \log 6\ \quad y \log 3= 3 \log 6$

$\longrightarrow \sf x = \dfrac{3 \log 6}{ \log 2}\ \quad y = \dfrac{3 \log 6}{ \log 3}$

$\longrightarrow \sf \purple{\dfrac{1}{x} = \dfrac{ \log 2}{ 3 \log 6}\ \; ; \quad \dfrac{1}{y} = \dfrac{ \log 3}{ 3 \log 6}}$

Our required value ,

$:\implies \sf \dfrac{1}{x}+\dfrac{1}{y}$

$:\implies \sf \dfrac{ \log 2}{3 \log 6}+\dfrac{\log 3}{3 \log 6}$

$:\implies \sf \dfrac{ \log 2 + \log 3}{3 \log 6}$

$\boxed{ \bullet\ \; \sf \red{\log a + \log b = \log ab}}$

$:\implies \sf \dfrac{ \log (2.3)}{3 \log 6}$

$:\implies \sf \dfrac{ \log 6}{3 \log 6}$

$:\implies \sf \dfrac{ 1 }{ 3 }\ \; \bigstar$

$\boxed{\bullet \; \; \sf \pink{ \dfrac{\textsf{\textbf{1}}}{\textsf{\textbf{x}}} + \dfrac{\textsf{\textbf{1}}}{\textsf{\textbf{y}}} = \dfrac{\textsf{\textbf{1}}}{\textsf{\textbf{3}}} }}$

★ ════════════════════ ★

Shortcut :

$\sf 2^x = 3^y = 216$

$\sf \to 2 = (216)^{\frac{1}{x}}\ \; \; \& \; \; \; 3 = (216)^{\frac{1}{y}}$

Multiply both the equations ,

$\to \sf 6 = (216)^{\frac{1}{x}+\frac{1}{y}}$

$\to \sf 6 = (6)^{3 \left( \frac{1}{x}+\frac{1}{y} \right) }$

$\sf \to 1 = 3 \left( \dfrac{1}{x} + \dfrac{1}{y} \right)$

$\sf \longrightarrow \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{3}$

Answer 2

Given :-

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

To Find :-

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

Solution :-

By Given we have ;

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

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 ;

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

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

Additional Information :-

• ${ \boxed { \tt { log ( a ) - log ( b ) = log \bigg ( \dfrac{a}{b} \bigg ) }}}$

• ${ \boxed { \tt log_{a} a = 1 }}$

• ${ \boxed { \tt log_{x} y = \dfrac{log_{e} y}{log_{e} x } }}$

• ${\boxed { \tt Natural \:\: log \:\: of \:\: x = ln ( x ) = log_{e} x }}$

• ${ \boxed { \tt { e = Euler's \:\: number }}}$

• ${ \boxed { \tt { e = \displaystyle \tt \lim_{ \tt n \to \infty } { \bigg ( \tt 1 + \dfrac{1}{n} \bigg )}^{n} }}}$

Maclaurin Series :-

• Binomial Expansion For all x² < 1

${ \boxed { \tt { \orange { {( 1 \pm x)}^{n} = 1 \pm \dfrac{nx}{1!} + \dfrac{n(n - 1)x²}{2!} + . . . . . . . }}}}$

${ \boxed { \tt { \red { {( 1 \pm x)}^{- n} = 1 \mp \dfrac{nx}{1!} + \dfrac{n(n + 1)x²}{2!} + . . . . . . . . }}}}$

• Maclaurin Series of Sin x :-

${ \boxed { \tt { \orange { Sin x = x - \dfrac{x³}{3!} + \dfrac{x⁵}{5!} - . . . . . . . . . . }}}}$

• Maclaurin Series of Cos x :-

${ \boxed { \tt { \green { Cos x = 1 - \dfrac{x²}{2!} + \dfrac{x⁴}{4!} - . . . . . . . . . . }}}}$

• Maclaurin Series of tan x :-

${ \boxed { \tt { \blue { tan x = x + \dfrac{x³}{3} + \dfrac{2x⁵}{15} + . . . . . . . . . . }}}}$

• Maclaurin Series of $\tt e^{x}$ :-

${ \boxed { \tt { \green { e^{x} = 1 + x + \dfrac{x²}{2!} + \dfrac{x³}{3!} + . . . . . . . . }}}}$

• Maclaurin Series of $\tt {tan}^{ - 1 } x$ If & only $\tt { |x| \leqslant 1 }$ :-

${ \boxed { \tt { \orange { {tan}^{-1} x = x - \dfrac{x³}{3} + \dfrac{x⁵}{5} - . . . . . . . . . . }}}}$