Question

Real numbers x , y satisfy
$\rm \green{2^x = 3^y = 216\ .\ Find\ \dfrac{1}{x}+\dfrac{1}{y}\ .}$
​

Answer 1

$\large{\dfrac{1}{3}}$

$\large\underline{\large\underline{\text{How I solved the problem}}}$

I took equal exponents on both sides to keep the equality. Then, I found the product to sum the exponent.

$\large\underline{\large\underline{\text{Explanation}}}$

The first equation

$\boxed{ \begin{aligned} &\rightarrow 216^{\frac{1}{x}}=(2^{x})^{\frac{1}{x}}\\\\&\rightarrow 216^{ \frac{1}{x} } =2^{ \frac{x}{x} } \\\\ &\rightarrow 216^{ \frac{1}{x} } = 2. \end{aligned} }$

The second equation

$\boxed{ \begin{aligned} &\rightarrow 216^{\frac{1}{y}}=(3^{y})^{\frac{1}{y}}\\\\&\rightarrow 216^{ \frac{1}{y} } =3^{ \frac{y}{y} } \\\\ &\rightarrow 216^{ \frac{1}{y} } = 3. \end{aligned} }$

Product of two equations

$\boxed{ \begin{aligned} &\rightarrow 216^{ \frac{1}{x} } \cdot 216^{ \frac{1}{y} }= 2 \cdot 3\\\\&\rightarrow 216^{( \frac{1}{x} + \frac{1}{y} )}=6 \text{ [ \because a^{x} \cdot a^{y} =a^{x+y} ] } \\\\&\rightarrow (6^{3})^{( \frac{1}{x} + \frac{1}{y} )} = 6 \text{ [ \because 216=6^{3} ]} \\\\&\rightarrow 6^{3( \frac{1}{x} + \frac{1}{y} )} = 6 \text{ [ \because (a^{x} )^{y} =a^{xy} ]} \\\\&\rightarrow 6^{( \frac{1}{x} + \frac{1}{y} )} = 6^{ \frac{1}{3} }. \end{aligned} }$

Hence,

$\cdots \longrightarrow \boxed{ \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{3} . } \ \underline{\small\text{Problem solved!}}$

Answer 2

Given :-

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

To Find :-

1/x + 1/y

Solution :-

$\sf 2^x=216[Given]$

On taking log of both LHS and RHS

$\sf log(2^x)=log(216)$

$\sf x \times log(2) = log(216)$

$\sf x=\dfrac{log(216)}{log(2)}$

On taking reciprocal of both side

$\sf \dfrac{1}{x}=\dfrac{log(2)}{log(216)} Eq.1$

Similarly

$\sf 3^y=216$

On taking log of both LHS and RHS

$\sf log(3^y)=log(216)$

$\sf log(3)\times y=log(216)$

$\sf y=\dfrac{log(216)}{log(3)}$

On taking reciprocal of both side

$\sf\dfrac{1}{y}=\dfrac{log(3)}{log(216)}Eq.2$

On adding Eq. 1 and Eq. 2

$\sf\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{log(2)}{log(216)}+\dfrac{log(3)}{log(216)}$

$\sf\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{log(2)+log(3)}{log(216)}$

$\sf\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{log(2\times3)}{log(216)}\bigg(log(a+b)=log(a\times b)\bigg)$

$\sf\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{log(6)}{log(216)}$

On prime factorizing 216

216 = 2 × 2 × 2 × 3 × 3 × 3

= 2³ × 3³

= (2 × 3)³$\sf a^m\times b^m=(ab)^m$

= (6)³

$\sf\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{log(6)}{log(6)^3}$

$\sf\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{log(6)}{3\;log(6)}$

On canceling log(6)

$\sf\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{\cancel{log(6)}}{3\cancel{log(6)}}$

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

Hence,

1/x + 1/y = 1/3