We'll call the result of the original function f and the argument g for sanity.
f = 2^{g-1} + 1
f-1 = 2^{g-1}
log2 (f-1) = g-1
g = 1 + log2 (f-1)
Not a, wrong sign on the -1 at the end.
f = 1/2 (ln(g/2) -1 )
2f = ln(g/2) -1
2f + 1 = ln(g/2)
g/2 = e^{2f + 1}
g = 2 e^{2f + 1}
b is a good pair of inverses, at least over the domain of the logarithm.
f = 4 ln(g^2)/e^2
fe^2 = 4 ln(g^2)
fe^2/4 = ln g^2
g^2 = e^{(fe^2)/4}
Now we take the square root so
[tex]g = \pm \sqrt{ e^{(fe^2)/4} }
[/tex]
The [tex]\pm[/tex] is telling us f is not one-to-one on its domain so doesn't have a true inverse.
f = 10^{g-10} - 10
f + 10 = 10^{g-10}
log(f+10) = g-10
g = 10 + log(f+10)
wrong sign again.
I'll go with choice B.
