Question

if 2^x=3^y=6^-z. prove that 1/x + 1/y + 1/z =0

Answer 1

Hi friend, Harish here.

Here is your answer:

Let:

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

Then:

[tex]2 = k^{ \frac{1}{x}} [/tex]

$3=k^{ \frac{1}{y}}$

$6 = k^{- \frac{1}{z}}$

We know that,

i) 3 × 2 = 6

ii) xᵃ × xᵇ = xᵃ⁺ᵇ


Then,

$3 \times 2 = 6$

Now, Substitute value of 3, 2,& 6.

$k^{ \frac{1}{x}} \times k ^{ \frac{1}{y}} = k^{- \frac{1}{z}}$

The bases are equal . 

So,

→ $\frac{1}{x} + \frac{1}{y} = - \frac{1}{z}$

→ $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$

Hence proved.
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Hope my answer is helpful to you.

Answer 2

Answer:

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