2007-01-02 05:33:13 · 7 answers · asked by Kat 1 in Science & Mathematics ➔ Chemistry
(t1/t2)=(a2/a1)^n
first take natural logs of
each side
ln(t1/t2)=ln(a2/a1)*n
n={ln(t1/t2)}/{ln(a2/a1)}
i hope that this helps
2007-01-02 06:24:49 · answer #1 · answered by Anonymous · 0⤊ 1⤋
Take the log of both sides and use properties of logarithms:
log(t1/t2) = log(a2/a1)^n
log(t1/t2) = n*log(a2/a1)
[log(t1/t2)] / [log(a2/a1)] = n
2007-01-02 13:39:54 · answer #2 · answered by lcamccandlj 3 · 2⤊ 0⤋
By using Logarithms
t1/t2 = (a2/a1)^n
Log (t1/t2) = log (a2/a1)^n
log t1 - log t2 = n(log a2 - log a1)
n = (log t1 - logt2) / (log a2 - log a1)
2007-01-02 16:38:05 · answer #3 · answered by lenpol7 7 · 0⤊ 2⤋
Homework a little hard?
Bloody idle stoooodents!
2007-01-02 13:38:20 · answer #4 · answered by Phil C 3 · 0⤊ 1⤋
Depends on what variable you are looking to solve for.
2007-01-02 13:40:34 · answer #5 · answered by Ruben Z 2 · 0⤊ 0⤋
Too many unknowns to get value of n.
2007-01-02 13:35:41 · answer #6 · answered by Anonymous · 1⤊ 2⤋
HUH? What catagory am I in?
2007-01-02 13:41:44 · answer #7 · answered by marisia 3 · 0⤊ 1⤋