solve for x... log2(x^2-1)-log2(x+1)=3?
回答 (7)
2009-12-31 5:52 am
✔ 最佳答案
Use the properties of logarithms:i.e, that log(a) - log(b) = log(a/b)
Then log2(x^2 -1) - log2(x+1) = 3 implies
log2(x^2-1)/(x+1) = 3
Then 2^3 = (x^2-1)/(x+1)
8 = (x+1)(x-1)/(x+1)
Then 8 = x-1 as the (x+1)'s cancel on the right hand side
==> Solution is x = 9
2016-12-18 1:53 am
(a million/3)^2 - x = 27 -x = 27 - (a million/3)^2 x = (a million/3)^2 - 27 in line with hazard you meant (a million/3)^(2 - x) = 27. consequently, (a million/3)^(2 - x) = 27 (a million/3)^(2 - x) = (a million/3)^(-3) 2 - x = -3 -x = -5 x = 5
2009-12-31 12:46 pm
log[2](x^2-1)-log[2](x+1)=3
x^2-1
---------=2^3
.x+1
.x^2-1=8(x+1)
x^2-1=8x+8
x^2-8x-1-8=0
x^2-8x-9=0
(x+1)(x-9)=0
x+1=0 x-9=0
x=-1 x=9
x=-1,9 answer//
x^2-1
---------=2^3
.x+1
.x^2-1=8(x+1)
x^2-1=8x+8
x^2-8x-1-8=0
x^2-8x-9=0
(x+1)(x-9)=0
x+1=0 x-9=0
x=-1 x=9
x=-1,9 answer//
2009-12-31 12:39 pm
x-1=8
x=9
God bless you.
x=9
God bless you.
2009-12-31 10:08 am
x=9
2009-12-31 6:42 am
log_2(x^2 - 1) - log_2(x + 1) = 3
log_2[(x^2 - 1^2)/(x + 1)] = 3
log_2{[(x + 1)(x - 1)]/(x + 1)} = 3
log_2(x - 1) = 3
x - 1 = 2^3
x = 8 + 1
x = 9
log_2[(x^2 - 1^2)/(x + 1)] = 3
log_2{[(x + 1)(x - 1)]/(x + 1)} = 3
log_2(x - 1) = 3
x - 1 = 2^3
x = 8 + 1
x = 9
2009-12-31 6:14 am
log2((x^2-1)/(x+1)=3
x-1=8
x=9
x-1=8
x=9
收錄日期: 2021-05-01 12:55:58
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20091230214659AA61kxM