Yucky logarithm and index questions? - index of babe
ok robbed in my records and indexes, but now we have a mission and a brain freeze, have two problems fascinates me is the solution for x (both values and 3-December exact locations of the appropriate answers)
7 ^ (2x) = 3 (3x +2)
and
log_2_ (x) less log_x_ (4) = 0
emphasized that I do on both sides of the base.
I know that my days of the law can not only explore ways to work with them, I get nowhere fast.
So please teach me to develop and Answers
Thanx babes xx
Friday, December 11, 2009
Index Of Babe Yucky Logarithm And Index Questions?
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3 comments:
7 ^ (2x) = 3 (3x +2)
log [7 ^ (2x)] = log [3 ^ (3x +2)]
log [(7 ^ 2) ^ x] = (3x + 2) log [3]
log [49 ^ x] = 3xlog [3] + 2 log [3]
xlog [49] = 3xlog [3] + 2 log [3]
Move all variables solveable [x] left:
xlog [49] - 3xlog [3] = 2 log [3]
x (log [49] - 3LOG [3]) = log 2 [3]
x = 2 log [3] / (log [49] - 3LOG [3])
x = log [3 ^ 2] / (log [49] - log [3 ^ 3])
x = log [9] / (log [49] - log [27])
x ≈ 3687
This is the base-10 I'm not sure if you want to base e (ln) or not.
7 ^ (2x) = 3 (3x +2)
(2x) IN7 = (3x +2) In3
[IN7 / In3] = (3x +2) / 2x = (3 / 2) + (1 / x)
1 / x = [IN7 / In3] - (3 / 2)
x = 1 / [(IN7 / In3) - (3 / 2)]
x = [] In3 / [IN7 - (3 / 2) In3]
-------------------------------------- ...
The solution of the problems solved
log_2_ (x) =- log_x_ (4)
[log x / log 2] = - [log4] logx (base for all registrations 2)
(log x) ^ 2 =- log4 / 2log2 =- log 2 / log 2 = -2
log x = √ 2 i (+/-)
x = 2 ^{(+/-) i √ 2)
x = 2 ^ (i √ 2) or 2 ^ (-i √ 2)
x = 2 ^ (i √ 2) or 1 / (2 ^ (i √ 2))
7 ^ (2x) = 3 (3x +2)
(2x) IN7 = (3x +2) In3
[IN7 / In3] = (3x +2) / 2x = (3 / 2) + (1 / x)
1 / x = [IN7 / In3] - (3 / 2)
x = 1 / [(IN7 / In3) - (3 / 2)]
x = [] In3 / [IN7 - (3 / 2) In3]
-------------------------------------- ...
The solution of the problems solved
log_2_ (x) =- log_x_ (4)
[log x / log 2] = - [log4] logx (base for all registrations 2)
(log x) ^ 2 =- log4 / 2log2 =- log 2 / log 2 = -2
log x = √ 2 i (+/-)
x = 2 ^{(+/-) i √ 2)
x = 2 ^ (i √ 2) or 2 ^ (-i √ 2)
x = 2 ^ (i √ 2) or 1 / (2 ^ (i √ 2))
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