Question

How do I solve Log {Log 12 + Log(base 2) (x + 1)} = 0

Answer 1

Since you designated only the third log expression as base 2, I will assume
the first two would be log(base 10), so
Log {Log 12 + Log(base 2) (x + 1)} = 0
10^0 = {Log 12 + Log(base 2) (x + 1)}
{Log 12 + Log(base 2) (x + 1)} = 1
Log 12 + Log(base 2) (x + 1) = 1
1.07918 + log(base 2) (x+1) = 1
log(base 2) (x+1) = -.07818
2^-.07818 = x+1
x+1 = .94659
x = appr -.0534

Even though I showed only a few decimal in my interim steps, on
my calculator I carried all available decimals, storing values in the calculator's memory.

If all logs are base 2, repeat the above steps with new values,
recalling that log₂ 12 = log 12/log 2