I'm going to assume the base of the logarithm was 10 (which it should be. If the base isn't clarified in the problem, automatically assume it's ten.)
antilog(15.6) = x (antilog is the same as log inverse, denoted as log^-1. I'll be using that to replace antilog since that's how I learned it. We are also trying to find x.)
log^-1(15.6) = x
Take the logarithm of both sides, as shown:
log[log^-1(15.6)] = log(x)
The log and log^-1 cancel out on the left-hand side, so we remain with:
log(x) = 15.6
Raise both sides of the equation as powers for 10.
10^log(x) = 10^15.6
Since the base of the logarithm on the left-hand side of the equation is 10, and it's being raised as an exponent of 10, they cancel out.
(For those who are confused, think of e and natural logs. e to the power of ln x is equal to x, because the exponent of e also has a natural log with a base of e. Your precalculus teachers would've gone over this.)
x = 10^15.6
Therefore, the antilog of 15.6 is 10^15.6.