Trying to make x the subject of this equation:?

Question

c = x^y/((x^y)+(b^y))

Answer 1

I dont know what you mean but i will try...

c = x^y/((x^y)+(b^y))

c((x^y)+(b^y))=x^y

cx^y+cb^y=x^y

cx^y-x^y= -cb^y

x^y(c-1)= -cb^y

x^y=(-cb^y)/(c-1)

x=y root of [(-cb^y)/(c-1)]

Answer 2

c = x^y/((x^y)+(b^y))

c = x^y/x^y  +  x^y/b^y

c = 1  + x^y/b^y

c - 1 = x^y/b^y

b^y(c - 1) = x^y

log b^y(c - 1) = log x^y

[log b^y(c - 1)] = y log x

[log b^y(c - 1)] / y =  log x

Antilog { [log b^y(c - 1)] / y} =  x

Answer 3

Divide top and bottom of the RHS by x^y:

c = 1/[1 + b^y/x^y] = 1/[1 + (b/x)^y]

c[1 + (b/x)^y] = 1

[1 + (b/x)^y] = 1/c

(b/x)^y = (1/c) - 1

ylog(b/x) = log[(1/c) - 1]

log(b/x) = (1/y)log[(1/c) - 1]

(b/x) = 10^{(1/y)log[(1/c) - 1]}

x = b*10^{(-1/y)log[(1/c) - 1]}

Check:

x^y = (b^y)10^{-log[(1/c) - 1]} = (b/y)[c/(1 - c)]

x^y/((x^y)+(b^y)) = [c/(1 - c)] / {[c/(1 - c)] + 1} = c / {c + 1 - c} = c

And this is correct