Question

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

Answer 1

Answer:

O

Step-by-step explanation:

Let 2^x=3^y=6^z=k

2^x=k

K^1/x=2

Similarly

k=3^y

K^1/y=3

Again

K=6^z

K^1/z=6=2*3

K^1/z=K^1/x*K^1/y

Equating base

1/z=1/x+1/y

1/x+1/y-1/z=0

Answer 2

$Answer \: \\ 2 {}^{x} = 3 {}^{y} = 6 {}^{z} \\ \\ let \: \: \: \: 2 {}^{x} = 3 {}^{y} = 6 {}^{z} = k \\ \\ 2 {}^{x} = k \: \: \: \: \: 3 {}^{y} = k \: \: \: \: \: and \: \: \: \: 6 {}^{z} = k \\ \\ 2 = k {}^{ \frac{1}{x} } \: \: \: ...Equation \: \: i \: \: \: \\ \\ 3 = k {}^{ \frac{1}{y} } \: \: \: ...Equation \: \: ii \: \\ \\ 6 = k {}^{ \frac{1}{z} } \: \: \: ...Equation \: iii \\ \\ 6 = k {}^{ \frac{1}{z} } \\ \\ 3 \times 2 = k {}^{ \frac{1}{z} } \\ \\ k {}^{ \frac{1}{x} } \times k {}^{ \frac{1}{y} } = k {}^{ \frac{1}{z} } \\ \\ k {}^{( \frac{1}{x} + \frac{1}{y})} = k {}^{ \frac{1}{z} } \\ \\ Now \: \: compare \: powers \: of \: k \\ \\ \frac{1}{x} + \frac{1}{y} = \frac{1}{z} \\ \\ \frac{1}{x} + \frac{1}{y} - \frac{1}{z} = 0 \: \: \: \: \: \: Hence \: \: proved \: : \: \:$