Question

prove that 1/x +1/y +1/z = 0
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Answer 1

QUESTION:

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

GIVEN:

• 2^x = 3^y = 6^{-z}

TO PROVE:

• 1/x +1/y +1/z = 0.

PROOF:

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

$\sf \leadsto Let \ 2^x = 3^y = 6^{-z} = k$

$\sf \leadsto \ 2^x = k$

$\boxed{ \large{ \gray{ \bold{If \ {x}^{m} = k,\ then \ x = k{^{1/m}} }}}}$

$\sf \leadsto \ 2 = k^{1/x}$

$\sf \leadsto 3^y = k$

$\sf \leadsto \ 3 = k^{1/y}$

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

$\sf \leadsto \ 6 = k^{1/ - z} = k^{ - 1/ z}$

$\boxed{ \large{ \gray{ \bold{2 \times 3 = 6}}}}$

Substitute the values of 2, 3 and 6.

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

$\boxed{ \large{ \gray{ \bold{x^m \times x^n = x^{m+n}}}}}$

$\sf \leadsto k^{(1/x) + (1/y)} = k^{ - 1/ z}$

$\boxed{ \large{ \gray{ \bold{If \ {x}^{m} = {x}^{n} ,\ then \ m = n }}}}$

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

$\bf Transpose \ - \dfrac{1}{ z} \ to \ the \ left$

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

HENCE PROVED.