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I have the following


I solved this by writing


so $k < 4,17$. The result should be $k > 4,17$, why is that? Where am I getting wrong?

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up vote 2 down vote accepted

Since $\log(0.46)$ is negative (because $0.46 \leq e$, or $0.46 \leq 10$, depending on what base we're using for the $\log$), when we divide through we must change the direction of the inequality. The rest looks fine.

The root of this matter is that $x \geq 0$ implies $-x \leq 0$. We can invent any example with concrete numbers: consider $-2 \leq 6$, and divide through by $-2$ to get $1 \geq -3$, not $1 \leq -3$. However, we could divide by $6$ and get $-\frac{1}{3} \leq 1$, still valid, because we're dividing by a positive number.

To answer your question in the comments:

When we have something like $\displaystyle \frac{2x + 3}{5x - 1} \geq 2$, we might be tempted to multiply by $5x-1$, but since you don't know its sign, it's better to avoid multiplying or dividing by a number which, for all we know, might be negative, and for example substract $2$ from both sides: $\displaystyle \frac{2x + 3}{5x - 1} - 2 = \frac{2x + 3 - 2(5x - 1)}{5x - 1} =\frac{-8x+5}{5x-1} \geq 0$, and keep the division as a last resource, when we'll have to use the next tip. To multiply or divide by an unknown quantity we have to consider every possible sign of what we're multiplying or dividing by, for example:

If $5x-1 \gt 0$, then we multiply without changing the direction: $\displaystyle 2x + 3 \geq 2(5x - 1)$, and if $5x-1 \lt 0$, then $\displaystyle 2x + 3 \leq 2(5x - 1)$; finally, we continue manipulating the expression to solve for $x$. In each case we must remember that we imposed a condition on $x$ (like $5x-1 \gt 0$ or $5x-1 \lt 0$), and take it into account at the end, even though it's not written on the equation.

The basic idea is:

  1. try to do simpler steps whenever possible; and
  2. we definitely have to change the direction if we're multiplying or dividing by a negative number, so if we don't know the sign, we must separate in cases.
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Thank you, what about a case when I have no numbers, let's say $k(s+1)<(s+2)$, can I write $k < \frac{(s+2)}{(s+1)}$ also if I don't know the sign of s+1 ? – Marco A. Jun 18 '12 at 18:10
$\log(0.46)$ is negative because $0.46 < 1$, not $e$ or $10$. – Logan Maingi Jun 18 '12 at 18:13
@Paul: That is precisely what you cannot do. If you don't know the sign of $s+1$, you have no way of knowing whether $k$ is greater or less than that quantity. (Look at what happens when you set $k=2, s=-2$ and what happens when you set $k=0, s=0$. Both of these satisfy your original inequality, but in one case $k < \frac{s+2}{s+1}$ and in the other case $k > \frac{s+2}{s+1}$.) – Micah Jun 18 '12 at 18:17
Thank you, I just realized it although it's been a long time that I've been messing around with inequalities. Equations like -2a < 3 are easy to solve but I need to stick what you just said in my head for other cases – Marco A. Jun 18 '12 at 18:30
@Paul: In regards to the $k(s+1)<(s+2)$ situation, the worst case scenario is that $s+1=0$. In that case, not only are we unable to divide by $s+1$, but $k$ could be literally any complex number and the inequality would still be satisfied! But the complex numbers cannot be ordered compatibly with the field operations, so we can't necessarily say $k<\mathrm{some\: number}$, even if we were given more information. – Cameron Buie Jun 18 '12 at 19:31

The log of $.46$ is negative, so when you divided, you should have changed the direction of the inequality.

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