Solve.I x-1l^[(lnlxl)^2 -lnx^2]=|x-1l^3.And next question :-) √log (5x-x^2)/4 with base 1/4.find x
Sigma Samantasinghar , 8 Years ago
Grade 11
Algebra> Solve.I x-1l^[(lnlxl)^2 -lnx^2]=|x-1l^3. ...
1 AnswersVikas TU
Last Activity: 8 Years ago
Again i am answering,
Case 1:
when x
U can equate Lnlxl)^2-Lnx^2 =3
or let logx = t
t^2 – 2t – 3 = 0
it has roots
t = -1 and t = 3
thus,
logx = -1 and logx = 3
and
x = 1/e
for x = e^3 solution rejected as e^3 is > 1.
Case 2: when x> 1
Again, t will have two roots as:
t = -1 and t = 3
x = 1/e and x = e^3
For x > 1 1/e is rejected and x = e^3 is accepted.
Now intersection for both cases dnot satisfies them simulatneously is something looks not utterabble.
But u can observe logically,
By intersection i mean to say:
one solution for one case is being get rejected for other case.
and same for vice versa.
Thus (Lnlxl^2)-Lnx^2 =3
cannot be equated.
Plz. post your next question as another post again!
Plz. post your next question as another post again!
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