SOLUTION: I have to solve N(t) =N_0(2)^t/d for t using logarithms.
This is what I have and I'd really appreicate some help on it.
logN(t) = log(N_0(2)^t/d)
logN(t) = t/d log(N_0)(2)
t/d
Algebra.Com
Question 29022: I have to solve N(t) =N_0(2)^t/d for t using logarithms.
This is what I have and I'd really appreicate some help on it.
logN(t) = log(N_0(2)^t/d)
logN(t) = t/d log(N_0)(2)
t/d = log(N_0(2)) / log(N(t)
t = log(N_0(2)) / log(N(t) * d
Is this right? Thank you.
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
SEE BELOW
I have to solve N(t) =N_0(2)^t/d for t using logarithms....WHAT IS THIS N_0(2) SYMBOL?..ANY WAY I AM IGNORING IT!BETTER PUT BRACKETS TO (T/D)AS OTHERWISE IT MAY MEAN ONLY^T AND MULTIPLY THE ANSWER WITH D....
This is what I have and I'd really appreicate some help on it.
logN(t) = log(N_0(2)^t/d)....OK
logN(t) = t/d log(N_0)(2)......OK
t/d =LOG(N(T))/ log(N_0(2))....NO..IT IS OTHERWAY ROUND...IT IS
t =D*{LOG(N(T))/ log(N_0(2))}.
IS THE CORRECT ANSWER...BUT CHECK IF N_0(2) HAS ANY MEANING.
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