Question

It is given that if ${{2}^{x}}={{3}^{y}}={{12}^{z}}$ , then we have to prove that $\dfrac{-2}{x}-\dfrac{1}{y}+\dfrac{1}{z}=0$ .

Answer 1

Hint: The given relation ${{2}^{x}}={{3}^{y}}={{12}^{z}}$ is in terms of powers of x, y and z. Therefore, we can take logarithm to convert the exponents as products and then use the logarithmic identities to convert the equations into the desired form.

Complete step-by-step answer:
In the question, it is given that ${{2}^{x}}={{3}^{y}}={{12}^{z}}$. As the terms are equal, let each term be equal to k.
${{2}^{x}}={{3}^{y}}={{12}^{z}}=k$……………..(1.1)
Now, we know the logarithmic identity that
$\log \left( {{a}^{b}} \right)=b\log \left( a \right).....................(1.2)$
Therefore, using equations (1.1) and (1.2), we can obtain 3 equations as
$\begin{align}
  & {{2}^{x}}=k \\
 & \Rightarrow \log \left( {{2}^{x}} \right)=\log k \\
 & \Rightarrow x\log \left( 2 \right)=\log k \\
 & \Rightarrow \log (2)=\dfrac{\log k}{x}.....................(1.3) \\
\end{align}$
For the second term, we get
$\begin{align}
  & {{3}^{y}}=k \\
 & \Rightarrow \log \left( {{3}^{y}} \right)=\log k \\
 & \Rightarrow y\log \left( 3 \right)=\log k \\
 & \Rightarrow \log (3)=\dfrac{\log k}{y}.....................(1.4) \\
\end{align}$
For the third term, we get
$\begin{align}
  & {{12}^{z}}=k \\
 & \Rightarrow \log \left( {{12}^{z}} \right)=\log k \\
 & \Rightarrow z\log \left( 12 \right)=\log k \\
 & \Rightarrow \log (12)=\dfrac{\log k}{z}.....................(1.5) \\
\end{align}$
Also, we know that the logarithm of product of two numbers is given by
$\log \left( ab \right)=\log a+\log b...................(1.6)$
Therefore, as $12=3\times 4$, we should have
$\log (12)=\log (3\times 4)=\log (3)+\log (4)..............(1.7)$
Also, we can write $4={{2}^{2}}$, therefore using the expression for logarithm of power from equation (1.2), we should have
$\log \left( 4 \right)=\log \left( {{2}^{2}} \right)=2\log \left( 2 \right)$
Using this in equation (1.7), we get
\[\log (12)=\log (12)=\log (3)+\log (4)=\log \left( 3 \right)+2\log \left( 2 \right)..............(1.8)\]
Now, we can replace log (12), log(3), log(2) from equations (1.3), (1.4) and (1.5), we get
\[\begin{align}
  & \log (12)=\log \left( 3 \right)+2\log \left( 2 \right) \\
 & \Rightarrow \dfrac{\log (k)}{z}=\dfrac{\log (k)}{y}+2\times \dfrac{\log (k)}{x} \\
\end{align}\]
Dividing $\log k$ on both sides, we obtain,
$\dfrac{\log k}{\left( \log k \right)\left( z \right)}=\dfrac{\log k}{\left( \log k \right)\left( y \right)}+2\dfrac{\log k}{\left( \log k \right)\left( x \right)}$
\[\begin{align}
  & \Rightarrow \dfrac{1}{z}=\dfrac{1}{y}+2\times \dfrac{1}{x} \\
 & \Rightarrow \dfrac{-2}{x}+\dfrac{-1}{y}+\dfrac{1}{z}=0 \\
\end{align}\]
From the above expression, we can say that we have proved what is asked.

Note: We should note that in the equations (1.3), (1.4) and (1.5), the value of k in all these equations should be the same because we had assumed the value of k from equation (1.1) and all the terms in the question were given to be equal.