Question

please look at the attachement and solve the 3 questions asked
plaese

Answer 1

1)loga- log b= log(a-b)
log a/b=log (a-b)              WKT loga/b=loga-logb
a/b=a-b
a=-b²/1-b⇒b²/b-1
option a is correct
 
2)f(x)=log(1-x)/(1+x)
then f(1-x)/(1+x)
f(1-x)/(1+x)⇒log(1/1  -  1-x/1+x)
                     (1/1  +1-x/1+x)
log(1+x-1+x)/(1+x+1-x)
log(2x/2)⇒logx
option a is correct

Answer 2

Log a / b = log (a - b)
  a/b = a - b
  a = ab - b²
  a (b-1) = b²
 a = b²/(b-1)
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f(x) = Log [ (1-x)/(1+x) ]
f( (1-x)/(1+x) ) = Log [  {1 - (1-x)/(1+x)} / { 1 + (1-x)/(1+x) } ]
   = Log [ 2x / 2   ]
   = Log x
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$x^{\frac{1}{3}}+y^{\frac{1}{3}}=-z^{\frac{1}{3}}\\\\cubing\\\\x+y+3(xy)^\frac{1}{3}(x^{\frac{1}{3}}+y^{\frac{1}{3}})=-z\\\\x+y+z=-3(xy)^\frac{1}{3}(-z)^{\frac{1}{3}}\\\\\frac{x+y+z}{3}=(xyz)^{\frac{1}{3}}\\\\answer=\frac{1}{3}Log(xyz)$