Question

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

Answer 1

Required Answer:-

Given:

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

To prove:

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

Proof:

Let us assume that,

$\sf \implies {2}^{x} = {3}^{y} = {6}^{ - z} = k \: (constant)$

So,

$\sf \implies {2}^{x} = k$

$\sf \implies 2 = {k}^{^{1}/_{x} } \: \: ....(i)$

And,

$\sf \implies {3}^{y} = k$

$\sf \implies 3= {k}^{^{1}/_{y} } \: \: ....(ii)$

Also,

$\sf \implies {6}^{ - z} = k$

$\sf \implies 6 = {k}^{^{1}/_{ - z} } \: \: ....(iii)$

Again,

$\sf \implies 6 = 2 \times 3$

From (i), (ii) and (iii), we can write that,

$\sf \implies {k}^{^{1}/_{ - z} } = {k}^{^{1}/_{x} } \times {k}^{^{1}/_{y}}$

$\sf \implies {k}^{^{1}/_{ - z} } = {k}^{^{1}/_{x} + ^{1}/_{y}}$

Comparing base, we get,

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

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

(Hence Proved)

Formula Used:

• $\sf x^{y} = x^{z} \implies y = z, x \neq 0$
• $\sf x^{y} = z \implies x = z^{^{1}/_{y}}$

Answer 2

$\huge\rm{Solution} \\ \\ \rm→ {2}^{x} = {3}^{y} = {6}^{-z} =K(Let) \\ \\ \rm→ {2}^{x} =K \: \: \: \: \: \: \: \: \: \: \: \: {3}^{y} =K \: \: \: \: \: \: \: \: \: \: \: \: {6}^{-z} =K \\ \rm→2=K \frac{1}{x}→(i) \: \: \: \: \: \: \: \: \: \: \: \: 3=K \frac{1}{y}→(ii) \: \: \: \: \: \: \: \: \: \: \: \: 6= \frac{1}{-z} →(iii) \\ \\ \rm \: We \: know \: that \\ \rm→2×3=6 \\ \rm→K \frac{1}{x} \times K \frac{1}{y} =K- \frac {1}{z} \\ \rm→K \frac{1}{x} + \frac{1}{y} =K \frac{ - 1}{z} \\ \rm→ \frac{1}{x} + \frac{1}{y} =- \frac{1}{z} \\ \rm→ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} =0$

$\huge\rm{Formula \: used,} \\ \\ \rm \: {a}^{n} = b \\ \rm[ {a}^{n} ] \: 1/n =b \: 1/n \\ \rm{\fbox {\: a=b \: 1/n}} \\ \\ \rm→ {a}^{m} \times {a}^{n} = {a}^{m + n}$

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