absolute value?

Question

If I take natural logs in an equation do I always need absolute value signs. And when I remove the logs does the absolute value always stay.

1. e^ydy = 1/x(dx)
gives e^y = ln|x| so does y = |x| (ignoring constant)

2. if I have x = y and I take ln of both sides

ln|y| = ln|x| then get rid of them.
|y| = |x| I appear to have changed the equation.
Is this wrong and why/why not? Thanks.

Answer 1

When you integrate something of the form du/u, the absolute value bars must be added every single time.



This is because the derivative of ln|x| is the same as ln(x), which is of course, 1/x. So when we integrate, we add the absolute value bars so that our original function stays defined for all x instead of x > 0.



When you're taking natural logs or any type of log for an equation, you do not add absolute value signs. Doing this alters the equation, which is a definite no-no since you're adding extraneous solutions.



For your first problem, when you integrate, you get:



e^y = ln|x| + C

y = ln(ln|x| + C)



y ≠ |x|. I think what you did was just make an error with solving for y. You need to "ln" both sides.



As for your second question, you can't make x and y have absolute values out of nowhere. What if x (also y) were negative in the first place? You're taking logs of negative values which is undefined. Making the variables have absolute values alters your equation making x and y carry only 0 or positive values.

Answer 2

which potential 2x-5 should be under 9 and additionally greater suitable than -9. So remedy 2x - 5 < 9 AND 2x - 5 > -9, via the strategies we've shown you in previous solutions. you should finally finally end up with x between -2 and seven, i.e. -2 < x < 7 be conscious emy is the only different answerer who's appropriate (up until finally now)

Answer 3

IN A LOG APPROACH, logN to base a =x,n is positive,a is positive,but a not=1,so to indicate n not as negative even in such orders we use absolute value.it is generally understood so no need to specify always.

Answer 4

Since log of negetive value is imaginary

absolute value is taken

Answer 5

i got no idea but think of it this way. you have a 50/50 chance of getting it right hope that helps