What is the solution?
What is the answer for:
8/(log_3 x^2) - 1/(log_x 3)=3
Note that log_3 x^2 means log that have base 3
2007-11-03 18:36:32 · 5 answers · asked by A A 1 in Science & Mathematics ➔ Mathematics
8/(2log _3 x) -- 1/log_x 3) =3
4/(log_3 x) --1/ (1/log_3 x) = 3
4/log_3 x -- log_3 x =3
Put Log_3 x = y
(4 --y^2)/y = 3
4 --y^2 =3y
Solving this quadriatic eqn, we get y=--4 or y=1
log_3 x = --4 gives x = 1/3^4 or 1/81
log_3 x = 1 gives x= 3 Thus x has two values
2007-11-03 19:13:05 · answer #1 · answered by Sundareswaran M 3 · 0⤊ 0⤋
3
2007-11-04 01:52:54 · answer #2 · answered by Anonymous · 0⤊ 0⤋
x=3
2007-11-04 02:48:45 · answer #3 · answered by none 2 · 0⤊ 0⤋
The answer is x=3
8/(log_3 x^2) - 1/(log_x 3)=3
can be written as
8/(2log_3 x) - 1/(log_x 3)=3
or
4/(log_3 x) - 1/(log_x 3)=3
Hence
(4-1)/(log_x 3)=3
Therefore
1/(log_x 3)=1
log_x 3=1
From the log definition
x^1=3
The answer is x=3
2007-11-04 01:51:08 · answer #4 · answered by iyiogrenci 6 · 0⤊ 0⤋
Greetings,
8/(2 log_3 x) - 1/(log_x 3) = 3
4/(log_3 x) - 1/(log_x 3) = 3
By inspection it looks like x=3 is a solution to this equation.
4 -1 = 3
Regards
2007-11-04 01:50:44 · answer #5 · answered by ubiquitous_phi 7 · 0⤊ 0⤋