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f(t) = 1000e^0.2. Find f^-1(p)
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I know this is a log problem
shuold there be a t in there somewhere?
oh, yeah
else it does not make sense
1000e^0.2t
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so f(x)= 1000e^(0.2t) y = 1000e^(0.2t) we need to find the inverse y/1000 = e^(0.2t) ln(y/1000) = ln(e^(0.2t)) ln(y/1000)=0.2tln(e) = 0.2t so t = ln(y/1000)/0.2
so when I solve the problem, do I substitute p in for y?
since it is f^-1(p)
@zzr0ck3r
before just used x and y
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yes
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