Question

if 2 ki power x is equal to 3 ki power Y is equal to 6 ki power minus Z then prove that 1 upon X + 1 upon y + 1 upon z =0.


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Answer 1

2
$2 ^{x} = 3 {}^{y }$
$3 {}^{y} = 6 { }^{ - z}$
$2 {}^{x} = k {}^{ \times \frac{1}{x} }$
$3 {}^{y} = k \times \frac{1}{y}$
$6 {}^{ - z} = k \times \frac{ - 1}{z}$
2
$2 {}^{x} \times 3 {}^{y} = 6 {}^{ - z}$
k
$k \times \frac{1}{x} \times k \times \frac{1}{y} = k \times \frac{ - 1}{z}$
$\frac{1}{x} + \frac{1}{y} = \frac{ - 1}{z}$
$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$

Answer 2

Answer:

2 ^{x} = 3 {}^{y } 2

x

=3

y

3 {}^{y} = 6 { }^{ - z} 3

y

=6

−z

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

x

=k

×

x

1

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

y

=k×

y

1

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

−z

=k×

z

−1

2

2 {}^{x} \times 3 {}^{y} = 6 {}^{ - z} 2

x

×3

y

=6

−z

k

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

x

1

×k×

y

1

=k×

z

−1

\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