log x^8/log x^2
x >0, x cant be1
2006-11-26 14:54:46 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
log {x^8} = 8 log x
log {x^2} = 2 log x
plug and chug answer is 4
2006-11-26 15:00:27 · answer #1 · answered by cw 3 · 0⤊ 0⤋
One rule you must know to solve this question, that is:
log (a) x^n = n log (a) x
The 'n' and 'x' is any variable number with the condition they are more than 0. 'a' is the base of a log. Such as log 2 base 5 = log (5) 2.
In your question, I could only assume that both log is base 10 as you did not mention anything.
log (10) x^8 / log (10) x^2
= 8 log (10) x / 2 log (10) x
= 8 / 2
= 4
For your information,
log (10) x / log (10) x = 1
All the best~!
2006-11-26 15:09:04 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Ra6f80db2e8075b1cf9f66682e4cc1acall made of basa6f80db2e8075b1cf9f66682e4cc1as is a6f80db2e8075b1cf9f66682e4cc1aqual of sum of logarithms: ln[a6f80db2e8075b1cf9f66682e4cc1aa6f80db2e8075b1cf9f66682e4cc1aa6f80db2e8075b1cf9f66682e4cc1aa6f80db2e8075b1cf9f66682e4cc1aa6f80db2e8075b1cf9f66682e4cc1aa6f80db2e8075b1cf9f66682e4cc1aa6f80db2e8075b1cf9f66682e4cc1aa6f80db2e8075b1cf9f66682e4cc1aa6f80db2e8075b1cf9f66682e4cc1aa6f80db2e8075b1cf9f66682e4cc1aa6f80db2e8075b1cf9f66682e4cc1aa6f80db2e8075b1cf9f66682e4cc1aa6f80db2e8075b1cf9f66682e4cc1aa6f80db2e8075b1cf9f66682e4cc1a] = lna6f80db2e8075b1cf9f66682e4cc1aa6f80db2e8075b1cf9f66682e4cc1aa6f80db2e8075b1cf9f66682e4cc1aa6f80db2e8075b1cf9f66682e4cc1aa6f80db2e8075b1cf9f66682e4cc1a + lna6f80db2e8075b1cf9f66682e4cc1aa6f80db2e8075b1cf9f66682e4cc1aa6f80db2e8075b1cf9f66682e4cc1aa6f80db2e8075b1cf9f66682e4cc1aa6f80db2e8075b1cf9f66682e4cc1a Tha6f80db2e8075b1cf9f66682e4cc1an ra6f80db2e8075b1cf9f66682e4cc1acall tha6f80db2e8075b1cf9f66682e4cc1a powa6f80db2e8075b1cf9f66682e4cc1ar you could ingredient out of logarithm: a6f80db2e8075b1cf9f66682e4cc1aa6f80db2e8075b1cf9f66682e4cc1aa6f80db2e8075b1cf9f66682e4cc1aa6f80db2e8075b1cf9f66682e4cc1aa6f80db2e8075b1cf9f66682e4cc1alna6f80db2e8075b1cf9f66682e4cc1a + a6f80db2e8075b1cf9f66682e4cc1aa6f80db2e8075b1cf9f66682e4cc1aa6f80db2e8075b1cf9f66682e4cc1aa6f80db2e8075b1cf9f66682e4cc1aa6f80db2e8075b1cf9f66682e4cc1alna6f80db2e8075b1cf9f66682e4cc1a and ultimately ra6f80db2e8075b1cf9f66682e4cc1acall that ln is a logarithm with a basa6f80db2e8075b1cf9f66682e4cc1a of (e^one million/2)(e^3/2) and tha6f80db2e8075b1cf9f66682e4cc1ara6f80db2e8075b1cf9f66682e4cc1afora6f80db2e8075b1cf9f66682e4cc1a lna6f80db2e8075b1cf9f66682e4cc1a = (e^one million/2)(e^3/2), tha6f80db2e8075b1cf9f66682e4cc1an: a6f80db2e8075b1cf9f66682e4cc1aa6f80db2e8075b1cf9f66682e4cc1aa6f80db2e8075b1cf9f66682e4cc1a + a6f80db2e8075b1cf9f66682e4cc1aa6f80db2e8075b1cf9f66682e4cc1aa6f80db2e8075b1cf9f66682e4cc1a = 4a6f80db2e8075b1cf9f66682e4cc1aa6f80db2e8075b1cf9f66682e4cc1a = (e^one million/2)(e^3/2)
2016-10-13 04:33:53 · answer #3 · answered by lipton 4 · 0⤊ 0⤋
log x^8/log x^2=8*log x /2*log x=4
2006-11-26 15:01:46 · answer #4 · answered by yupchagee 7 · 0⤊ 0⤋
x^8 can be written as x^(2 * 4), (x^2)^4
Now, as log x^a is equivalent to a * log x
log x^8 = log (x^2)^4 = 4 log x^2
Thus, log x^8/log x^2 = 4 * (log x^2/log x^2) = 4
2006-11-27 20:25:46 · answer #5 · answered by pravkas 2 · 0⤊ 0⤋
log (x^8)/log (x^2) = (8 log x)/(2 log x) = 4
2006-11-26 15:16:00 · answer #6 · answered by Pascal 7 · 0⤊ 0⤋
i did this a while ago (like 3 yeaars) either take the log function of both side and it would reduce into somethin simple or..... thats all i remember sry
2006-11-26 14:57:26 · answer #7 · answered by Anonymous · 0⤊ 0⤋