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The number of value(s) of `x` satisfying `1-log_(3)(x+1)^(2)=1/2log_(sqrt(3))((x+5)/(x+3))` is

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`1-log(3^(2))(x+1)^(2)=1/2log_(3^(1//2))((x+5)/(x+3))`
`=1-2/2log_(3)|x+1|2/2=log_(3)((x+5)/(x+3))`
`=log_(3)(3/(|x+1|))=log_(3)((x+5)/(x+3))`
`implies3/(|x+1|)=(x+5)/(x+3)`
Case I1 `x+1gt0impliesxgt-1`
`implies3(x+3)=(x+1)(x+5)`
`impliesx^(2)+3x-4=0`
`impliesx=-4` or `x=1`
`x=-4` rejected `(xgt-1)`
`:.x=1`
Case II `x+1lt0impliesxlt-1`
`3(x+3)-1(x+1)(x+5)`
`impliesx^(2)+9x+14=0`
`impliesx=-2` or `x=-7`
`:.` Set of value of `x` `={-7, -2, 1}`
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