If g(x)=2log(x-3)+1 and the base is 10 what is g-1(x)?

We have g(x) = 2 log (x - 3) +
1

Let y = g(x) = 2 log (x - 3) +
1

=> y - 1 = 2 log (x -
3)

=> (y - 1)/2 = log (x -
3)

taking the base of the log as
10

=> 10^[(y - 1)/2] = x -
3

=> x = 10^[(y - 1)/2] +
3

interchange x and
y

=> y = 10^[(x - 1)/2] +
3

Therefore the inverse function of g(x)
is

g^-1(x) = y = 10^[(x - 1)/2] +
3