Answer:
uhhhhhhhhhhhhhhhhh
Step-by-step explanation:
i wish i knew
Given that
log (x+y)/5 =( 1/2) {log x+logy}
We know that
log a+ log b = log ab
⇛log (x+y)/5 =( 1/2) log(xy)
We know that log a^m = m log a
⇛log (x+y)/5 = log (xy)^1/2
⇛log (x+y)/5 = log√(xy)
⇛(x+y)/5 = √(xy)
On squaring both sides then
⇛{ (x+y)/5}^2 = {√(xy)}^2
⇛(x+y)^2/5^2 = xy
⇛(x^2+y^2+2xy)/25 = xy
⇛x^2+y^2+2xy = 25xy
⇛x^2+y^2 = 25xy-2xy
⇛x^2+y^2 = 23xy
⇛( x^2+y^2)/xy = 23
⇛(x^2/xy) +(y^2/xy) = 23
⇛{(x×x)/xy} +{(y×y)/xy} = 23
⇛(x/y)+(y/x) = 23
Therefore, (x/y)+(y/x) = 23
Hence, the value of (x/y)+(y/x) is 23.
I think it is -160 I’m not sure tho
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