I think both ways are correct, why did I get such different answers then?

[Loki123]

I tried to solve it two different ways and actually got different results. I can't pinpoint a mistake. So what's going on?

 

[Deleted member 4993]

I tried to solve it two different ways and actually got different results. I can't pinpoint a mistake. So what's going on? View attachment 30414

Nice presentation!

However in the second line you made a mistake.

If x > y then (-x) < (-y)

 

[Harry_the_cat]

In solution II, the (x+5) needs an index of -1 also in the first line.

 

[JeffM]

My goodness. So many places to go wrong. Tricky problem.

Method 1

$$- \{ log_7(x) + log_7(x + 5) \} \ge - 1 \implies log_7(x) + log_7(x + 5) \le 7.$$
Multiplying by a negative number reverses an inequality.

It is true that

$$x^2 + 5x - 7 = 0 \implies x \le \dfrac{-5 - \sqrt{53}}{2} < 0 \text { or } x \le \dfrac{-5 + \sqrt{53}}{2}.$$
And from that, one can deduce

$$x^2 + 5x - 7 \le 0 \implies x \in \left (- \infty,\ - \dfrac{-5 - \sqrt{53}}{2} \ \right ] \bigcup \left [\ 0, \ \dfrac{\sqrt{53} - 5}{2} \ \right ].$$
But we started with $log_7(x)$. Can x be less than or equal to 0?

Thus we get

$$0 < x \le \dfrac{\sqrt{53} - 5}{2}.$$
Method II

$$- log_7(x) - log_7(x + 5) \ge - 1 \implies log_7 \left ( \dfrac{1}{x} \right ) + log_7 \left ( \dfrac{1}{x + 5} \right ) \ge log_7 \left ( \dfrac{1}{7} \right ) \implies$$
$$log_7 \left ( \dfrac{1*1}{x*(x +5)} \right ) \ge log_7 \left ( \dfrac{1}{7} \right ) \implies log_7 \left ( \dfrac{1}{x^2 + 5x} \right ) \ge log_7 \left ( \dfrac{1}{7} \right ) \implies \dfrac{1}{x^2 + 5x} \ge \dfrac{1}{7}.$$
Does cross multiplying in this case reverse the inequality? No. Why not in this specific case?

So we get to

$$7 \ge x^2 + 5x \implies x^2 + 5x \le 7 \implies x^2 + 5x - 7 \le 0.$$
The two methods (if done without error) do converge. But cross multiplication in inequalities is super-tricky. I try to avoid it.

 

[Loki123]

Nice presentation!

However in the second line you made a mistake.

If x > y then (-x) < (-y)

oh yeah, i missed that. However, even when I correct that the two results do not match.

 

[Loki123]

My goodness. So many places to go wrong. Tricky problem.

Method 1

$$- \{ log_7(x) + log_7(x + 5) \} \ge - 1 \implies log_7(x) + log_7(x + 5) \le 7.$$
Multiplying by a negative number reverses an inequality.

It is true that

$$x^2 + 5x - 7 = 0 \implies x \le \dfrac{-5 - \sqrt{53}}{2} < 0 \text { or } x \le \dfrac{-5 + \sqrt{53}}{2}.$$
And from that, one can deduce

$$x^2 + 5x - 7 \le 0 \implies x \in \left (- \infty,\ - \dfrac{-5 - \sqrt{53}}{2} \ \right ] \bigcup \left [\ 0, \ \dfrac{\sqrt{53} - 5}{2} \ \right ].$$
But we started with $log_7(x)$. Can x be less than or equal to 0?

Thus we get

$$0 < x \le \dfrac{\sqrt{53} - 5}{2}.$$
Method II

$$- log_7(x) - log_7(x + 5) \ge - 1 \implies log_7 \left ( \dfrac{1}{x} \right ) + log_7 \left ( \dfrac{1}{x + 5} \right ) \ge log_7 \left ( \dfrac{1}{7} \right ) \implies$$
$$log_7 \left ( \dfrac{1*1}{x*(x +5)} \right ) \ge log_7 \left ( \dfrac{1}{7} \right ) \implies log_7 \left ( \dfrac{1}{x^2 + 5x} \right ) \ge log_7 \left ( \dfrac{1}{7} \right ) \implies \dfrac{1}{x^2 + 5x} \ge \dfrac{1}{7}.$$
Does cross multiplying in this case reverse the inequality? No. Why not in this specific case?

So we get to

$$7 \ge x^2 + 5x \implies x^2 + 5x \le 7 \implies x^2 + 5x - 7 \le 0.$$
The two methods (if done without error) do converge. But cross multiplication in inequalities is super-tricky. I try to avoid it.

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.

 

[JeffM]

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.

$$-log_7(x) - log_7(x + 5) \ge - 1 \implies (-1)log_7(x) - log_7(x + 5) \ge (-1)log_7(7) \implies$$
$$log_7 \left ( \dfrac{1}{x} \right ) - log_7(x + 5) \ge log_7 \left ( \dfrac{1}{7} \right )\implies log_7 \left ( \dfrac{\dfrac{1}{x}}{x + 5} \right ) \ge log_7 \left ( \dfrac{1}{7} \right ) \implies$$
$$log_7 \left ( \dfrac{\dfrac{1}{x}}{\dfrac{x + 5}{1}}\right ) \ge log_7 \left ( \dfrac{1}{7} \right ) \implies log_7 \left ( \dfrac{1}{x} * \dfrac{1}{x + 5} \right ) \ge log_7 \left ( \dfrac{1}{7} \right ) \implies$$
$$log_7 \left ( \dfrac{1}{x^2 + 5x} \right ) \ge log_7 \left ( \dfrac{1}{7} \right ) \implies \dfrac{1}{x^2 + 5x} \ge \dfrac{1}{7}.$$
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.

 

[Loki123]

Oh you could do that. Again though you must do it correctly.

$$-log_7(x) - log_7(x + 5) \ge - 1 \implies (-1)log_7(x) - log_7(x + 5) \ge (-1)log_7(7) \implies$$
$$log_7 \left ( \dfrac{1}{x} \right ) - log_7(x + 5) \ge log_7 \left ( \dfrac{1}{7} \right )\implies log_7 \left ( \dfrac{\dfrac{1}{x}}{x + 5} \right ) \ge log_7 \left ( \dfrac{1}{7} \right ) \implies$$
$$log_7 \left ( \dfrac{\dfrac{1}{x}}{\dfrac{x + 5}{1}}\right ) \ge log_7 \left ( \dfrac{1}{7} \right ) \implies log_7 \left ( \dfrac{1}{x} * \dfrac{1}{x + 5} \right ) \ge log_7 \left ( \dfrac{1}{7} \right ) \implies$$
$$log_7 \left ( \dfrac{1}{x^2 + 5x} \right ) \ge log_7 \left ( \dfrac{1}{7} \right ) \implies \dfrac{1}{x^2 + 5x} \ge \dfrac{1}{7}.$$
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.