log5(x^2-1/x-1)
then what from here???
2007-06-17 22:50:21 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
What from there? I wont do the whole problem for you, since that is how you are wording your question.
Realize that x^2 - 1 is the difference between two squares.
2007-06-17 22:55:03 · answer #1 · answered by Anonymous · 0⤊ 1⤋
what do you mean by the box?? -?
log5(x^2 - 1^2) - log 5 (x-1)
=log5(x-1)(x+1)-log5 (x-1)
=log5{[(x-1)(x+1)] / (x-1)}
=log5(x+1)
can this be accepted?
2007-06-17 23:06:10 · answer #2 · answered by student 1 · 0⤊ 0⤋
=
log5 [(x+1)(x-1)]/x-1
log5 (x+1)
5^1=x
x=5
2007-06-17 22:56:46 · answer #3 · answered by Anonymous · 0⤊ 1⤋
log5 (x^2-1/x-1)
= log5 [(x+1)(x-1)]/(x-1)
=log5(x+1)
2007-06-17 23:00:21 · answer #4 · answered by wen 1 · 0⤊ 1⤋
Let log mean log base 5 in the following:-
log [(x - 1).(x + 1)] - log (x - 1)
= log [(x - 1).(x + 1) / (x - 1) ]
= log (x + 1)
If x = 4 this becomes :-
log (5) = 1 (as an example)
2007-06-18 04:49:17 · answer #5 · answered by Como 7 · 0⤊ 0⤋
Sorry but your first line is wrong. It should be
log[5(x² - 1) / 5(x-1)]
= log [(x² - 1) / (x-1)]
= log [(x - 1)(x + 1) / (x - 1)]
= log (x + 1) .................... end
You can't evaluate it, you can only simplify it, there is no equation or value of x
Just realised that you may be working in logs base 5 . Please ignore first and second lines...rest still holds
2007-06-17 23:05:27 · answer #6 · answered by fred 5 · 0⤊ 0⤋
log5(x^2-1/x-1)
log5[(x+1)(x-1)/(x-1)]
log5(x+1)
2007-06-17 23:03:34 · answer #7 · answered by harry m 6 · 0⤊ 1⤋