Question

Prove that x+y+6 =0​

Answer 1

Answer:

Here is your answer:

Let:

2^{x} = 3^{y} = 6^{-z} =k2

x

=3

y

=6

−z

=k

Then:

2 = k^{ \frac{1}{x}}2=k

x

1

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

y

1

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

−

z

1

We know that,

i) 3 × 2 = 6

ii) xᵃ × xᵇ = xᵃ⁺ᵇ

Then,

3 \times 2 = 63×2=6

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

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

x

1

×k

y

1

=k

−

z

1

The bases are equal .

So,

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

x

1

+

y

1

=−

z

1

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

x

1

+

y

1

+

z

1

=0

Hence proved.

_____________________________________________-

Hope my answer is helpful to you.

Answer 2

Answer:

your answer is in photo