Question

If ${2^x} = {3^y} = {6^{ - z}}$, then what is the value of $\left( {\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}} \right)$?
$
  {\text{A}}{\text{. 0}} \\
  {\text{B}}{\text{. 1}} \\
  {\text{C}}{\text{. }}\dfrac{3}{2} \\
  {\text{D}}{\text{. }} - \dfrac{1}{2} \\
 $

Answer 1

Hint: Here, we will proceed by assuming the given equation equal to k. Then, we will represent each side of this equation in terms of k and we will use the general formulas of natural logarithmic function which are $\ln \left( {{a^b}} \right) = b\ln \left( a \right)$ and $\ln \left( a \right) + \ln \left( b \right) = \ln \left( {a \times b} \right)$.

Complete step-by-step answer:

Given, ${2^x} = {3^y} = {6^{ - z}}$

Let us suppose that each term in the above given equation is equal to k i.e., ${2^x} = {3^y} = {6^{ - z}} = k{\text{ }} \to (1{\text{)}}$

By taking ln on all the sides of equation (1), we have

$\ln \left( {{2^x}} \right) = \ln \left( {{3^y}} \right) = \ln \left( {{6^{ - z}}} \right) = \ln \left( k \right)$

Using the formula $\ln \left( {{a^b}} \right) = b\ln \left( a \right)$ in the above equation, we get

$x\ln \left( 2 \right) = y\ln \left( 3 \right) = \left( { - z} \right)\ln \left( 6 \right) = \ln \left( k \right){\text{ }} \to {\text{(2)}}$

From equation (2), we can write

\[

   \Rightarrow x\ln \left( 2 \right) = \ln \left( k \right) \\

   \Rightarrow x = \dfrac{{\ln \left( k \right)}}{{\ln \left( 2 \right)}}{\text{ }} \to {\text{(3)}}

\\

 \]

From equation (2), we can write

\[

   \Rightarrow y\ln \left( 3 \right) = \ln \left( k \right) \\

   \Rightarrow y = \dfrac{{\ln \left( k \right)}}{{\ln \left( 3 \right)}}{\text{ }} \to {\text{(4)}}
\\

 \]

From equation (2), we can write

\[

   \Rightarrow \left( { - z} \right)\ln \left( 6 \right) = \ln \left( k \right) \\

   \Rightarrow - z = \dfrac{{\ln \left( k \right)}}{{\ln \left( 6 \right)}} \\

   \Rightarrow \left( { - z} \right)\ln \left( 6 \right) = \ln \left( k \right) \\

   \Rightarrow z = - \dfrac{{\ln \left( k \right)}}{{\ln \left( 6 \right)}}{\text{ }} \to {\text{(5)}}

\\

 \]

By substituting the values of x, y and z from equations (3), (4) and (5) respectively in the expression $\left( {\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}} \right)$, we get

$

  \left( {\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}} \right) = \dfrac{1}{{\left[ {\dfrac{{\ln \left( k \right)}}{{\ln \left( 2 \right)}}} \right]}} + \dfrac{1}{{\left[ {\dfrac{{\ln \left( k \right)}}{{\ln \left( 3 \right)}}} \right]}} + \dfrac{1}{{\left[ { - \dfrac{{\ln \left( k \right)}}{{\ln \left( 6 \right)}}} \right]}} \\

   \Rightarrow \left( {\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}} \right) = \dfrac{{\ln \left( 2 \right)}}{{\ln \left( k \right)}} + \dfrac{{\ln \left( 3 \right)}}{{\ln \left( k \right)}} - \dfrac{{\ln \left( 6 \right)}}{{\ln \left( k \right)}} \\

 $

By taking ln(k) as the LCM of all the terms on the RHS of the above equation, we get

 $

   \Rightarrow \left( {\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}} \right) = \dfrac{{\ln \left( 2 \right) + \ln \left( 3 \right) - \ln \left( 6 \right)}}{{\ln \left( k \right)}} \\

   \Rightarrow \left( {\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}} \right) = \dfrac{{\left[ {\ln \left( 2 \right) + \ln \left( 3 \right)} \right] - \ln \left( 6 \right)}}{{\ln \left( k \right)}} \\

 $

Using the formula $\ln \left( a \right) + \ln \left( b \right) = \ln \left( {a \times b} \right)$ in the above equation, we get

$

   \Rightarrow \left( {\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}} \right) = \dfrac{{\left[ {\ln \left( {2 \times 3} \right)} \right] - \ln \left( 6 \right)}}{{\ln \left( k \right)}} \\

   \Rightarrow \left( {\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}} \right) = \dfrac{{\ln \left( 6 \right) - \ln \left( 6 \right)}}{{\ln \left( k \right)}} \\

   \Rightarrow \left( {\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}} \right) = \dfrac{0}{{\ln \left( k \right)}} \\

   \Rightarrow \left( {\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}} \right) = 0 \\

 $

Therefore, the value of the expression $\left( {\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}} \right)$ is 0.

Hence, option A is correct.

Note: In this particular problem, we have found the values of x, y and z in terms of k (which is assumed) and then these values are substituted in the expression $\left( {\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}} \right)$ and then we will simplify this expression in terms of k by taking the lowest common method (LCM) which comes out to be equal to 0.