Question

$\: if \: \: 2^x = 3^y = 4^z , then \: \: \: prove \: \: \: that \: \: \: 1/x \: + \: \: 1/y \: \: - \: \: 1/z = 0 \: \: or \: \: z = xy/x+y$
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Answer 1

$\huge\fcolorbox{blue}{green}{LET}$

• 2^x = 3^y = 6^z = k

• 2^x = k [ 2 = k^1/x ]

• 3^y = k [ 3 = k^1/y ]

• 6^z = k [ 6 = k^1/z]

$\huge\fcolorbox{blue}{green}{Now}$

2 * 3 = 6 [ (k)^1/x * (k)^1/y = (k)^1/z

(k)^1/x + 1/y = (k)^1/z

ON EQUATING THE POWER FROM BOTH SIDE:

• 1/x + 1/y = 1/z

• 1/x + 1/y - 1/z = 0 [PROVED]

• 1/z = 1/x + 1/y

• 1/z = x + y / xy

• z = xy / x + y [PROVED]