Question

If log (x-y/2)=1/2 (log x + log y - log z), then find the value of x/y + y/x

Answer 1

log(x-y/2)=log((xy/z)^{1/2})
compare
x-y/2=(xy/z)^{1/2}

Answer 2

㏒ {$\frac{x-y}{2}$} = $\frac{1}{2}$ (㏒ x + ㏒ y - ㏒ z)
2㏒ {$\frac{x-y}{2}$} = ㏒ x + ㏒ y - ㏒ z
㏒ $\frac{x-y}{2})^{2}$ = ㏒ $\frac{xy}{z}$

Remove ㏒ on both sides,

$( \frac{x-y}{2})^{2}$ = $\frac{xy}{z}$
(x²+y²-2xy)/4 = xy/z
(x²+y²-2xy)/xy = 4/z
x²/xy + y²/xy - 2xy/xy = 4/z
x/y + y/x - 2 = 4/z
x/y + y/x = 4/z + 2
x/y + y/x = (4+2z)/z

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If your question is,
If log (x-y/2)=1/2 (log x + log y), then find the value of x/y + y/x
Then this is the answer.

㏒ {$\frac{x-y}{2}$} = $\frac{1}{2}$ (㏒ x + ㏒ y)
2㏒ {$\frac{x-y}{2}$} = ㏒ x + ㏒ y
㏒ $\frac{x-y}{2})^{2}$ = ㏒ xy

Remove ㏒ on both sides,

$( \frac{x-y}{2})^{2}$ = xy
(x²+y²-2xy)/4 = xy
(x²+y²-2xy)/xy = 4
x²/xy + y²/xy - 2xy/xy = 4
x/y + y/x - 2 = 4
x/y + y/x = 4+2
x/y + y/x = 6

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Hope it helps