Given that $x, y, a$ are real numbers that satisfy $(\log_ax)^2+(\log_ay)^2 - \log_a(xy)^2 \le 2 $ and $\log_ay\ge 1$, find the range of $\log_ax^2y$

Question

Given that $x, y, a$ are real numbers that satisfy

$$(\log_ax)^2+(\log_ay)^2 - \log_a(xy)^2 \leq 2 \text{ and } \log_ay\geq1$$

Find the range of $\log_ax^2y$

My try: Let $b= \log_ax, c= \log_yx, 2b+c = \log_ax^2y$

$$b^2 + c^2 - 2b - 2c -2 \leq0$$

$$1- \sqrt{-c^2+2c+3}\leq b\leq 1+ \sqrt{-c^2+2c+3}$$

$$2(1- \sqrt{-c^2+2c+3})+c\leq 2b+c\leq 2(1+ \sqrt{-c^2+2c+3})+c$$

substituting $c = 1$

$$-1\leq 2b+c\leq7$$

but my teacher said that the answer is $3+2\sqrt{5}$ instead of $7$

Does $\log_a(xy)^2$ mean $(\log_a (xy))^2$ or $\log_a((xy)^2) \text{ ?}$ Either of those two would be unambiguous. $\qquad$ — Michael Hardy, Sep 20 '20 at 14:07

score 1 · Accepted Answer · answered Sep 20 '20 at 13:36

1

Proceeding from $$b^2 + c^2 - 2b - 2c -2 \leq 0 \\ \Rightarrow (b-1)^2+(c-1)^2 \le 2^2$$ which is a disk centered at $(1,1)$ with radius $2$.

Now, $2b+c=p$ is a family of parallel lines in the plane. Range of $p$ will be given by tangent lines to the disk. There will be two tangents, each at distance $2$ from center $(1,1)$.

You can now use the distance formula of a point from a line.

answered Sep 20 '20 at 13:36

cosmo5

10,629

You also have to use your condition $log_{a}y = b \ge1$ which should give your final answer. – cosmo5 Sep 20 '20 at 14:51

Given that $x, y, a$ are real numbers that satisfy $(\log_ax)^2+(\log_ay)^2 - \log_a(xy)^2 \le 2 $ and $\log_ay\ge 1$, find the range of $\log_ax^2y$

1 Answers1