i'm having trouble... help?
2007-12-03 19:41:05 · 9 answers · asked by Hans Lambert 1 in Science & Mathematics ➔ Mathematics
x+1/x-1= 10
x+1= 10x-10
11= 9x
x=11/9
2007-12-03 21:49:49 · answer #1 · answered by gonpatrick21 3 · 1⤊ 0⤋
There is a rule for logarithm functions.
1. log (a*b) = log (a) + log (b)
similarily,
2. log (a/b) = log (a) - log (b)
Your problem:
log (x + 1) - log(x - 1) = 1
1. Use rule #2 above!
log ( (x + 1) / (x - 1) ) = 1
2. Raise to the ten power to cancel out the log.
Remember: 10^log(a) = a!
(x + 1) / (x - 1) = 10
(x + 1) = 10 (x - 1)
x + 1 = 10x - 10
9x - 11 = 0
x = 11/9
Enjoy!
2007-12-03 19:49:53 · answer #2 · answered by kromped 2 · 2⤊ 0⤋
Take logs as log to base 10:-
log [ (x + 1) / (x - 1) ] = 1
(x + 1) / (x - 1) = 10^1
x + 1 = 10x - 10
9x = 11
x = 11/9
2007-12-03 20:08:31 · answer #3 · answered by Como 7 · 2⤊ 0⤋
log(x+1) - log(x-1)=1 .......... Supposing log base is 10.
This is a difference of log. Hence -
=> Log [ (x+1) divided by (x-1) ] = 1
=> Log [ (x+1) / (x-1) ] = log 10 ... since log 10 = 1
Taking anti log of both the sides we get -
(x+1) / (x-1) = 10
ie x + 1 = 10 ( x - 1 )
ie x +1 = 10 - x
ie 11 = 9x
ie x = 11/9................. Answer
2007-12-03 20:07:44 · answer #4 · answered by Pramod Kumar 7 · 1⤊ 1⤋
log(x+1) - log(x-1) = 1
LHS = log(x+1) - log(x-1)
= log[(x+1)/(x-1)]
assuming its log to the base 10
since 10^2 = 100 and log100 = 2 then..
10^1 = (x+1) /( x-1)
10 = (x+1) / (x-1)
10x - 10 = x + 1
9x = 11
x = 11 / 9
2007-12-03 19:48:51 · answer #5 · answered by sleepy 2 · 1⤊ 0⤋
log(x+1) -- log(x--1) = 1
=> log{(x+1)/(x--1)} = log(10)
=> 10(x -- 1) = (x + 1)
=> 9x = 11
=> x = 11/9
2007-12-03 20:19:06 · answer #6 · answered by Anonymous · 1⤊ 0⤋
log(x+1) - log(x-1)=1
Use the secong log law
log a -log b = log(a/b)
So
log(x+1) - log(x-1)=1
log ( (x+1)/(x-1)) =1
Use the definition of a log
log a (baseb) =c if b^c =a
so
log ( (x+1)/(x-1)) =1
(x+1)/(x-1) =10^1
(x+1)/(x-1) =10
x+1 = 10x-10
9x=11
x=11/9
Check that it is a valid solution.
Yes
2007-12-03 19:49:25 · answer #7 · answered by Anonymous · 1⤊ 0⤋
log(x+1) - log(x-1)=1
log[ (x+1) /(x-1) ]=1 . . . . anti log both sides
[ (x+1) /(x-1) ] = 10
(x+1) = 10(x-1)
x+1 = 10x-10
9x = 11
x = 11/9
2007-12-03 20:01:21 · answer #8 · answered by CPUcate 6 · 1⤊ 0⤋
ok uh [log(x) = log10(x)]
log(x+1) - log(x-1) = 1
log[(x+1)/(x-1)] = 1
=> log10[(x+1)/(x-1)] = 1
loga(x) = y <=> y = a^x
(x+1)/(x-1) = 10^1
(x+1)/(x-1) = 10
x+1 = 10 (x-1)
x+1 = 10x - 10
11 = 9x
x = 11/9
[(and i checked on my calc, x=11/9 is correct) or it works for this eqn]
2007-12-03 20:01:40 · answer #9 · answered by desert_storm_494 2 · 1⤊ 0⤋