err no idea.... for that one part....
and for another question how would i do....
ln x^4_3 square root 7x-2 over (x+5)^2
the _ that separates the 4 and 3 is just a space, so disregard anything else
these 2 questions are stumping me!!
and for question #2 "Write as a sum and/or difference of logs, simplify when possible"
p.s. you guys are awesome for helping me!
its 2:09 A.M. and i have a huge math quiz tommorow; my math teacher did not help what so ever!!
anywaysss...
log(x+4)=2log(x-2) math question?
2009-03-18 10:09 am
回答 (3)
2009-03-18 11:16 am
✔ 最佳答案
log(x + 4) = 2log(x - 2)log(x + 4) = log[(x - 2)^2]
x + 4 = (x - 2)(x - 2)
x + 4 = x^2 - 4x + 4
x^2 - 4x - x - 4 + 4 = 0
x^2 - 5x = 0
x(x - 5) = 0
x = 0
x - 5 = 0
x = 5
∴ x = 0 , 5
2009-03-18 10:46 am
Question 1
log(x + 4) = log (x - 2)²
x + 4 = (x - 2)²
x + 4 = x² - 4x + 4
x² - 5x = 0
x (x - 5) = 0
x = 0 , x = 5
Question 2
Sorry but unable to decipher what you have given.
log(x + 4) = log (x - 2)²
x + 4 = (x - 2)²
x + 4 = x² - 4x + 4
x² - 5x = 0
x (x - 5) = 0
x = 0 , x = 5
Question 2
Sorry but unable to decipher what you have given.
2009-03-18 10:37 am
For the first one:
log(x+4) = 2log(x-2)
and 2log(x-2) = log((x-2)^2) (this a log rule: aLog(b) = Log(b^a)
so: x+4 = (x-2)^2
x+4 = x^2 - 4x + 4
x^2 - 5x = 0
x(x - 5) = 0
so x = 0; OR x = 5
(don't have time for the second one atm)
log(x+4) = 2log(x-2)
and 2log(x-2) = log((x-2)^2) (this a log rule: aLog(b) = Log(b^a)
so: x+4 = (x-2)^2
x+4 = x^2 - 4x + 4
x^2 - 5x = 0
x(x - 5) = 0
so x = 0; OR x = 5
(don't have time for the second one atm)
收錄日期: 2021-05-01 12:11:41
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090318020951AAda0jT