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#103
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N. Korean citizen (07/17/2012) [-]
Isn't this supposed to be solved with a logarithmic function?
f(x) = 2^x - 2
2 = 2^x
We know that b^x = y and log(base b)y = x
so
log(base 2)X = 2
then you do something like change of base, or something with the 2.
and then solve from there (I got lazy)
f(x) = 2^x - 2
2 = 2^x
We know that b^x = y and log(base b)y = x
so
log(base 2)X = 2
then you do something like change of base, or something with the 2.
and then solve from there (I got lazy)