how do I solve
Logbase5(x+1) - Logbase5(x-1) = 2
2006-10-11 17:21:35 · 6 answers · asked by ctjarig 1 in Science & Mathematics ➔ Mathematics
log base5 (x + 1) - log base 5 ( x - 1) = 2
using log(a / b) = log a - log b
log base 5 ((x + 1) / (x - 1)) = 2
=> 5 ^ 2 = ((x + 1) / (x - 1))
=> 25 = ((x + 1) / (x - 1))
=> 25(x - 1) = x + 1
=> 25x - 25 = x + 1
=> 24x = 26
=> x = 26 / 24
=> x = 13 / 12
Hope it helps :)
2006-10-11 17:29:20 · answer #1 · answered by fsm 3 · 0⤊ 0⤋
therefore logbase5 [(x+1)/(x-1)] = 2
therefore [(x+1)/(x-1)] = 5^2 = 25
therefore (x+1) = 25(x-1)
Solve for x: x = 13/12
2006-10-12 00:27:27 · answer #2 · answered by Morkeleb 3 · 0⤊ 0⤋
Logbase5(x+1) - Logbase5(x-1) = 2
Logbase5[(x+1)/(x-1)] = 2
(x+1)/(x-1) = 5^2
x + 1 = 25x - 25
26 = 24x
x = 13/12
Th
2006-10-12 01:08:04 · answer #3 · answered by Thermo 6 · 0⤊ 0⤋
Logbase5((x+1)/(x-1)) = 2
(x+1)/(x-1) = 25
x + 1 = 25x - 25
24x = 26
x = 13/12
2006-10-12 00:35:24 · answer #4 · answered by Helmut 7 · 0⤊ 0⤋
logbase5(x+1) - logbase5(x-1) = 2
or, logbase5[(x+1)/(x-1)] = 2
or, (x+1)/(x-1) = 5^2
or, x+1 = 25x - 25
or, 24x = 26
Therefore, x= 26/24 = 1.08333
2006-10-12 00:27:47 · answer #5 · answered by Innocence Redefined 5 · 0⤊ 0⤋
x = 13/12 is the correct answer
2006-10-12 01:04:53 · answer #6 · answered by shamu 2 · 0⤊ 0⤋