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Mathematics
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Logarithms
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ln(x+z) + ln(x-2y+z) =2ln(x-z) then, (a)2y=x+z (b)y^2=xz (c)z^2=yz (d)y=\[\frac{ 2xy }{ x+z }\]
mmm maybe this could work ? ln(x+z) + ln(x-2y+z) = ln ([x+z] *[x-2y+z]) = ln (x-z) ^2
Probably , cancel out those logs wait doing it
yx+yz-2zx = 0 after that
x^2-2xy+xz+xz-2yz+z^2 = x^2 - 2zx + z^2
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-2xy+xz+xz-2yz = - 2zx -2y(x+z)+2xz=-2xz -2y(x+z) =-4xz y(x+z) =2xz mm getting this again :-\ \(y=\large \frac{2xz}{x+z}\) sure from the last option ? can u check if you dint made a typo ?
Yes you are right i might have made a arithmatic error thanks a million
ohhh np ;)
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