Question

if 2 power x =3 power y = 6 power -z ,prove that 1/x+1/y+1/z = 0

Answer 1

If  $2^{x} = 3^{y}  = 6^{-z}$ then $\frac{1}{x} + \frac{1}{y} +\frac{1}{z} = 0$ is true.

• Let us assume $2^{x} = 3^{y}  = 6^{-z} = a$
• now $2^{x} = a$
• $a^{\frac{1}{x} } = 2$
• Similarly , $a^{\frac{1}{y} } = 3$
• and $a^{\frac{1}{-z} } = 6$
• Now we start by simplifying $a^{\frac{1}{-z} } = 6$
• $a^{\frac{1}{-z} }$= 3 × 2
• but $a^{\frac{1}{x} } = 2$ and $a^{\frac{1}{y} } = 3$
• Therefore, $a^{\frac{1}{-z} }$ =$a^{\frac{1}{x}}$ × $a^{\frac{1}{y}}$
• $a^{\frac{1}{-z} } = a^{\frac{1}{x} + \frac{1}{y} }$
• As the base on both sides are same we compare powers and equate them
• $\frac{1}{-z} = \frac{1}{x} + \frac{1}{y}$
• $\frac{1}{x} + \frac{1}{y} +\frac{1}{z} = 0$
• Hence proved

Answer 2

Answer:

If   then  is true.

Let us assume

now

Similarly ,

and

Now we start by simplifying

= 3 × 2

but  and

Therefore,  = ×

As the base on both sides are same we compare powers and equate them

Hence proved