Question

if 2^x = 3^y = 6^z , prove that 1/x + 1/y - 1/z = 0 or z = (x-y)/(x+y)

Answer 1

$\mathbb{ SOLUTION }$ :

$let ,\\ {2}^{x} = {3}^{y} = {6}^{z} = k \\ \\ = > {2}^{x} = k \: \: , \: \: {3}^{y} = k \: \: and \: \: {6}^{z} = k \\ \\ = > 2 = {(k)}^{ \frac{1}{x} } \: \: , \: \: 3 = {(k)}^{ \frac{1}{y} } \: \: and \: 6 = {(k)}^{ \frac{1}{z} } \\ \\ now. \\ \\ 6 = 2 \times 3 \\ \\ = > {(k)}^{ \frac{1}{z} } = {(k)}^{ \frac{1}{x} } \times {(k)}^{ \frac{1}{y} } \\ \\ now \: \: equating \: \: their \: \: powers \\ we \: \: get \\ \\ \frac{1}{x} + \frac{1}{y} = \frac{1}{z} \\ \\ so \\ \\ \frac{1}{x} + \frac{1}{y} - \frac{1}{z} = 0 \: \: \: \ \\ \\ now \\ \\ \frac{1}{z } = \frac{1}{x} + \frac{1}{y} \\ \\ = > \frac{1}{z} = \frac{x + y}{xy} \\ \\ = > z = \frac{xy}{x + y}$

Hence, Proved

Answer 2

HEY MATE HERE IS UR ANSWER

LET 2^y=3^y=6^z = K

TERE FOR

2^ Y = K

2= K ^ 1/X. --------( 1)

SIMILARLY 3= K 1/Y AND 6 = K^1/Z --------(2)

NOW WE KNOW THAT 2 X 3 = 6 ------------- 3)

THERE FOR

2 X 3 = 6 AND WHICH IS EQUAL TO

K ^ 1/X * K ^1/Y = K ^1 /Z

= K ^(1/X + 1/Y) = K^ 1/Z

= 1/X + 1/Y = 1/Z

= 1/X + 1/Y -- 1/Z = 0

HENCE PROVED IF YOU FOUND IT HELPFUL PLEASE MARK AS BRAINLIEST


SOLUTION :

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