5^21 x 4^11=2x10^n
what is the value of n?
How do you solve this?
2007-12-28 09:02:37 · 5 answers · asked by tlbarron 1 in Science & Mathematics ➔ Mathematics
Take the log of both sides
21*log(5) + 11Log(4) = Log(2) + nlog(10)
21*log(5) + 11Log(4) - Log(2) = nlog(10)
n = [21*log(5) + 11Log(4) - Log(2)] / log(10)
n = [21*(0.69897) + 11*(0.60206) - (0.30103)] / 1
n= 14.67837 + 6.6266 - 0.30103
n = 21.00394
2007-12-28 09:20:32 · answer #1 · answered by WhatWasThatNameAgain? 5 · 0⤊ 0⤋
OK
5^21 x 4^11=2x10^n ; the trick is to see that 4^11 = 2^22, so
5^21 x 2^22 = 2x10^n
(5^21)(2^21)(2) = (2)(10^n) ; now divide both sides by 2
5^21* 2^21 = 10^n
10^21 = 10^n
n = 21
Hope that helps.
2007-12-28 17:24:03 · answer #2 · answered by pyz01 7 · 0⤊ 0⤋
5^21 x 4^11 = 2 x 10^n
Or 2x10^n = 5^21 x 2^21 x 2
Or 10^n = 10^21
Or n = 21
2007-12-28 17:17:55 · answer #3 · answered by sv 7 · 0⤊ 0⤋
5^21 * 4^11 = 2 * 10^n
5^21 * (2^2)^11 = 2 * 10^n
5^21 * 2^22 = 2 * 10^n
5^21 * 2^21 * 2 = 2 * 10^n
10^21 * 2 = 2 * 10^n
10^21 = 10^n
Therefore, n = 21
2007-12-28 17:18:40 · answer #4 · answered by falzoon 7 · 0⤊ 0⤋
21log 5 + 11log4 = log2 + nlog10
n = 21log5 + 11log4 - log2
= 21
2007-12-28 17:17:36 · answer #5 · answered by norman 7 · 0⤊ 0⤋