By taking logarithms, find the smallest possible integer, n, for which 4^n < 5^(n-1)
Ive been puzzled over this for ages. Any help is really appreciated!
2007-11-02 14:26:04 · 5 answers · asked by Stevie B 2 in Science & Mathematics ➔ Mathematics
4^n < 5^(n-1)
taking logs
n log 4 < (n-1) log 5
0.602 n < 0.7( n - 1)
0.602n < 0.7 n - 0.7
subtract 0.602 n from both sides
0< 0.1 n - 0.7
add 0.7 both sides
0.7 < 0.1n
divide by 0.1
0.7/0.1 < n
n>7/1
so smallest integer that satisfies the given condition is 8
2007-11-02 14:54:03 · answer #1 · answered by mohanrao d 7 · 0⤊ 0⤋
Try a sneaky approach, take the log base 4 of both sides, then
n < n-1 log4 (5) which is appx n= (7/6)*(n-1)
You should be able to get a value of n for this.
2007-11-02 14:35:14 · answer #2 · answered by cattbarf 7 · 0⤊ 0⤋
Take ln of both sides:-
n ln 4 < (n - 1) ln 5
1.39 n < (n - 1) (1.61)
1.39 n < 1.61 n - 1.61
1.61 < 0.22 n
n > 7.32
Smallest integer is 8.
2007-11-02 21:41:38 · answer #3 · answered by Como 7 · 0⤊ 1⤋
I took the logs n set fire to them
2007-11-02 15:14:57 · answer #4 · answered by Luv 2 Larf 2 · 0⤊ 0⤋
ln4^n=ln5^(n-1)
nln4=(n-1)ln5
nln4=nln5-ln5
nln4-nln5=-ln5
n(ln4-ln5)=-ln5
n=-ln5/(ln4-ln5)
n=ln5/(ln5-ln4)
n=7.2
2007-11-02 14:33:17 · answer #5 · answered by someone else 7 · 1⤊ 0⤋