Question

2x=3y=6z then prove that 1/x+1/y+1/z=0

Answer 1

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Answer 2

GIVEN :

2^x = 3^y = 6^-z

TO FIND :

PROVE: 1/x + 1/y + 1/z = 0

SOLUTION:

We can simply solve the problem as under :

Let us suppose,

2^x = 3^y = 6^-z = K

so,

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

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

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

we know that,

2 × 3 = 6

Putting the values of 2, 3 and 6; we get ;

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

or,

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

SINCE, K IS NOT EQUAL TO 0

So,

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

or,

1/x + 1/y + 1/z = 0 HENCE PROVED.

CORRECT QUESTION:

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

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