Question

hey matte.. pls answer it..❤️❤️❤️❤️❤️

Answer 1

We know that $2^{x} = 3^{y} = 6^{-z}$

Now, let's take a constant as 'C'

Alright, we can write:

$2^{x} = C \\3^{y} = C \\6^{-z} = C$

Looking that that, we can know that:

$2 = C^{\dfrac{1}{x} }\\3 = C^{\dfrac{1}{y} }\\6 = C^{\dfrac{1}{-z} }$

Also, we know that 2 × 3 = 6

Hence, we can write this as:

$C^{\dfrac{1}{x} }*C^{\dfrac{1}{y} } = C^{\dfrac{1}{-z} }$

Looking at that, we can know that:

$\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{-1}{z}$

If we know add 1/z on both sides, we get:

$\dfrac{1}{x}+\dfrac{x}{y}+\dfrac{1}{z}=0$

∴Hence proved