I think both ways are correct, why did I get such different answers then?
- Thread starter Loki123
- Start date
D
Deleted member 4993
Guest
Nice presentation!
However in the second line you made a mistake.
If x > y then (-x) < (-y)
Harry_the_cat
Elite Member
- Joined
- Mar 16, 2016
- Messages
- 3,693
In solution II, the (x+5) needs an index of -1 also in the first line.
My goodness. So many places to go wrong. Tricky problem.
Method 1
[math]- \{ log_7(x) + log_7(x + 5) \} \ge - 1 \implies log_7(x) + log_7(x + 5) \le 7.[/math]
Multiplying by a negative number reverses an inequality.
It is true that
[math]x^2 + 5x - 7 = 0 \implies x \le \dfrac{-5 - \sqrt{53}}{2} < 0 \text { or } x \le \dfrac{-5 + \sqrt{53}}{2}.[/math]
And from that, one can deduce
[math]x^2 + 5x - 7 \le 0 \implies x \in \left (- \infty,\ - \dfrac{-5 - \sqrt{53}}{2} \ \right ] \bigcup \left [\ 0, \ \dfrac{\sqrt{53} - 5}{2} \ \right ].[/math]
But we started with [imath]log_7(x)[/imath]. Can x be less than or equal to 0?
Thus we get
[math]0 < x \le \dfrac{\sqrt{53} - 5}{2}.[/math]
Method II
[math]- log_7(x) - log_7(x + 5) \ge - 1 \implies log_7 \left ( \dfrac{1}{x} \right ) + log_7 \left ( \dfrac{1}{x + 5} \right ) \ge log_7 \left ( \dfrac{1}{7} \right ) \implies[/math]
[math]log_7 \left ( \dfrac{1*1}{x*(x +5)} \right ) \ge log_7 \left ( \dfrac{1}{7} \right ) \implies log_7 \left ( \dfrac{1}{x^2 + 5x} \right ) \ge log_7 \left ( \dfrac{1}{7} \right ) \implies \dfrac{1}{x^2 + 5x} \ge \dfrac{1}{7}.[/math]
Does cross multiplying in this case reverse the inequality? No. Why not in this specific case?
So we get to
[math]7 \ge x^2 + 5x \implies x^2 + 5x \le 7 \implies x^2 + 5x - 7 \le 0.[/math]
The two methods (if done without error) do converge. But cross multiplication in inequalities is super-tricky. I try to avoid it.
Method 1
[math]- \{ log_7(x) + log_7(x + 5) \} \ge - 1 \implies log_7(x) + log_7(x + 5) \le 7.[/math]
Multiplying by a negative number reverses an inequality.
It is true that
[math]x^2 + 5x - 7 = 0 \implies x \le \dfrac{-5 - \sqrt{53}}{2} < 0 \text { or } x \le \dfrac{-5 + \sqrt{53}}{2}.[/math]
And from that, one can deduce
[math]x^2 + 5x - 7 \le 0 \implies x \in \left (- \infty,\ - \dfrac{-5 - \sqrt{53}}{2} \ \right ] \bigcup \left [\ 0, \ \dfrac{\sqrt{53} - 5}{2} \ \right ].[/math]
But we started with [imath]log_7(x)[/imath]. Can x be less than or equal to 0?
Thus we get
[math]0 < x \le \dfrac{\sqrt{53} - 5}{2}.[/math]
Method II
[math]- log_7(x) - log_7(x + 5) \ge - 1 \implies log_7 \left ( \dfrac{1}{x} \right ) + log_7 \left ( \dfrac{1}{x + 5} \right ) \ge log_7 \left ( \dfrac{1}{7} \right ) \implies[/math]
[math]log_7 \left ( \dfrac{1*1}{x*(x +5)} \right ) \ge log_7 \left ( \dfrac{1}{7} \right ) \implies log_7 \left ( \dfrac{1}{x^2 + 5x} \right ) \ge log_7 \left ( \dfrac{1}{7} \right ) \implies \dfrac{1}{x^2 + 5x} \ge \dfrac{1}{7}.[/math]
Does cross multiplying in this case reverse the inequality? No. Why not in this specific case?
So we get to
[math]7 \ge x^2 + 5x \implies x^2 + 5x \le 7 \implies x^2 + 5x - 7 \le 0.[/math]
The two methods (if done without error) do converge. But cross multiplication in inequalities is super-tricky. I try to avoid it.
Last edited:
oh yeah, i missed that. However, even when I correct that the two results do not match.
I understand the process. However what I don't understand is why I can't only give index -1 to x and solve by formula loga - logb.
Oh you could do that. Again though you must do it correctly.
[math]-log_7(x) - log_7(x + 5) \ge - 1 \implies (-1)log_7(x) - log_7(x + 5) \ge (-1)log_7(7) \implies[/math]
[math]log_7 \left ( \dfrac{1}{x} \right ) - log_7(x + 5) \ge log_7 \left ( \dfrac{1}{7} \right )\implies log_7 \left ( \dfrac{\dfrac{1}{x}}{x + 5} \right ) \ge log_7 \left ( \dfrac{1}{7} \right ) \implies[/math]
[math]log_7 \left ( \dfrac{\dfrac{1}{x}}{\dfrac{x + 5}{1}}\right ) \ge log_7 \left ( \dfrac{1}{7} \right ) \implies log_7 \left ( \dfrac{1}{x} * \dfrac{1}{x + 5} \right ) \ge log_7 \left ( \dfrac{1}{7} \right ) \implies[/math]
[math]log_7 \left ( \dfrac{1}{x^2 + 5x} \right ) \ge log_7 \left ( \dfrac{1}{7} \right ) \implies \dfrac{1}{x^2 + 5x} \ge \dfrac{1}{7}.[/math]
But what you cannot do is to just change a minus sign into a plus sign as you did in line 1 of your method II.
Thank youu, I can't believe I didn't see that.