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@Zer0kx: Go
Yes, using math :D
log_b(x) = log_a(x) / log_a(b)
In other words, logarithm functions differ only by a constant factor
@Zer0kx: Go
Yes, using math :D
log_b(x) = log_a(x) / log_a(b)
In other words, logarithm functions differ only by a constant factor