Question

If ${2^x} = {3^3} = {6^{ - z}}$ , prove that $\left( {\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 0} \right)$

Answer 1

Hint: In this particular type of question firstly we have to take the value of the expression as k and rearrange in the form if k to the power n. Then use substitution to get to the desired answer.

Complete step-by-step answer:
Let ${2^x} = {3^y} = {6^{ - z}} = k$
Then,
2 = ${k^{\dfrac{1}{x}}}$
3 = ${k^{\dfrac{1}{y}}}$
6 = ${k^{ - \dfrac{1}{z}}}$
We know that,
$3 \times 2 = 6$
Substituting value of 2 , 3 and 6 from above
${k^{\dfrac{1}{x}}} \times {k^{\dfrac{1}{y}}} = {k^{ - \dfrac{1}{z}}}$
Since , ${x^a} \times {x^b} = {x^{a + b}}$
${k^{\dfrac{1}{x}}} \times {k^{\dfrac{1}{y}}} = {k^{\dfrac{1}{x} + \dfrac{1}{y}}} = {k^{\left\{ { - \dfrac{1}{z}} \right\}}}$
$ \Rightarrow \dfrac{1}{x} + \dfrac{1}{y} = - \dfrac{1}{z}$ ( since bases are equal )
$ \Rightarrow \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 0$

Note: Remember to recall the basic exponential identities to make the question easier to solve . Note that in this question we have used an indirect way to solve the question , to prove $\left( {\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 0} \right)$ , we first proved that ${k^{\dfrac{1}{x}}} \times {k^{\dfrac{1}{y}}} = {k^{ - \dfrac{1}{z}}}$ and as their bases are common we used the exponential identity to prove them equal. Remember this method whenever you deal with such a type of question.