Question

Q. If 2^x = 3^y = 6^z, then prove that 1/x + 1/y - 1/z = 0.​

Answer 1

ANSWER:

Given:

• 2^x = 3^y = 6^z

To Prove:

• 1/x + 1/y - 1/z = 0

Proof:

$\text{We are given that,}\\\\:\longrightarrow2^x=3^y=6^z\\\\\text{Let, these be equal to a constant k, i.e.}\\\\:\implies2^x=3^y=6^z=k\\\\:\implies2^x=k\implies2=k^{\frac{1}{x}}- - - -(1)\\\\:\implies3^y=k\implies3=k^{\frac{1}{y}}- - - -(2)\\\\:\implies6^z=k\implies6=k^{\frac{1}{z}}- - - -(3)\\\\\text{Multiplying (1) & (2),}\\\\:\implies2\times3=k^{\frac{1}{x}}\times k^{\frac{1}{y}}\\\\:\implies6=k^{\frac{1}{x}}\times k^{\frac{1}{y}}\\\\\text{From (3),}\\\\:\implies k^{\frac{1}{z}}=k^{\frac{1}{x}}\times k^{\frac{1}{y}}\\\\\text{As, bases are same, we compare the powers,}\\\\:\implies\dfrac{1}{z}=\dfrac{1}{x}+\dfrac{1}{y}\\\\:\implies\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{z}\\\\\text{Transposing RHS to LHS,}\\\\\bf{:\implies\dfrac{1}{x}+\dfrac{1}{y}-\dfrac{1}{z}=0 = RHS}\\\\\text{\bf{Hence Proved!!}}$

Answer 2

Question

$If 2^x = 3^y = 6^z, \: then \: prove \: that \: 1/x + 1/y - 1/z = 0.$

Here is your answer:

Let:

$2^{x} = 3^{y} = 6^{-z} =k2x=3y=6−z=k</p><p>$

Then:

$2 = k^{ \frac{1}{x}}2=kx1</p><p>$

$3=k^{ \frac{1}{y}}3=ky1$

$</p><p>6 = k^{- \frac{1}{z}}6=k−z1$

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.

$k^{ \frac{1}{x}} \times k ^{ \frac{1}{y}} = k^{- \frac{1}{z}}kx1×ky1=k−z1$

The bases are equal . 

So,

→

$\frac{1}{x} + \frac{1}{y} = - \frac{1}{z}x1+y1=−z1</p><p>$

→ 

$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0x1+y1+z1=0$