The following question about the application of logarithmic function properties is one of the questions in my course exam. For this question, our teacher only gave the answer without giving the detailed dissolution steps, so I added my own solution steps, but I am not sure whether it is correct, so please help me point out the mistakes in my dissolution steps and give me the correct approach.
Given the function $f(x)=| \log_e (x) |$, if $m>n>0$ and $f(m)=f(n)$, select the correct terms from the following conclusions:
$ \ \ a.m+n=2; \qquad b.mn=1; \qquad c.\frac{2}{m} + \frac{1}{n} \ge 2\sqrt 2; \qquad d.\frac{1}{m} + \frac{2}{n} \ge 3; $
For option a, I don't know how to get the specific value of $m+n $by using the question conditions, so I skip the steps of a here. If you can help me supplement the steps of option a, I will be very grateful to you.
For choice b, use the two conditions of the question:
- a. $f(m)=f(n)$
- b. $m>n>0$
The following conclusions can be drawn:
- a. $\|log_em\| = \|log_en\|$, $ log_em = -log_en$
- b. $m>1$,
- c. $ 0<n<1 $,
- d. According to a. the conclusion can be drawn :$m*n=1,\quad m=\frac{1}{n}$
- For option c, using the conclusion $m=\frac{1}{n}$from 2, make the following inferences:
$$ 2n + \frac{1}{n} \ge 2\sqrt2 \\ \Rightarrow 2n^2 + 1 \ge 2\sqrt2 *n \\ \Rightarrow 2n^2 - 2\sqrt2 *n + 1 \ge 0 \\ \Rightarrow (\sqrt2n - 1)^2 \ge 0 $$
This converts the inequality of option c into a quadratic equation, let $f(x)=(\sqrt2x-1)^2$, from the graph of the function, which has only one point of intersection with the X-axis: $(\frac{\sqrt2}{2},0) $, the function image is above the X-axis, which satisfies $(\ sqrt2n-1)^2 \ge 0$, so option B is also the correct conclusion.
- For option d, the following inference is made using the conclusion $m=\frac{1}{n}$obtained from 2:
$$ \frac{1}{m} + \frac{2}{n} \ge 3 \\ \Rightarrow n^2 - 3n + 2 \ge 0 $$
Let $f(x)= n^ 2-3n + 2$, according to the quadratic equation to find the root formula, we can see that the equation has two real roots, respectively $x_1=1,x_2=2$, we can see from the function image, when $1 \le x \le 2$, $f(x) \le 0$is constant. According to the conclusion of 2. c.$0<n<1 $, $f(x) is always greater than 0$in the interval (0,1). So $n^2-3n + 2 \ge 0$works. So choice D is also the correct conclusion.
To sum up, the conclusion of this topic is BCD.
Please help me check if there is any problem with my steps and correct them. I would be very grateful if you could help me supplement the analysis steps of option a.
As far as I can tell, you've shown that Options B,C and D are right.
For Option A, note that $f(2)=f\left(\frac{1}{2}\right).$ However, $2+\frac{1}{2}≠2.$ Option A isn't right.