| | | | | | | | | | | | | | | | | | | | | | | | | | | | | |
#103 - 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)
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)
User avatar #114 to #103 - ithinkimfunny (07/17/2012) [-]
yeah, for f(x) = 0
User avatar #113 to #103 - mrblueftw (07/17/2012) [-]
why is it that when theres a math joke on FJ all the smart people come out
User avatar #115 to #113 - ithinkimfunny (07/17/2012) [-]
but it's not a smart comment lol
User avatar #108 to #103 - sidedoor (07/17/2012) [-]
No sir.
 Friends (0)