AIME 2017 II · 第 7 题

Solution 1

Note the equation $\log(kx)=2\log(x+2)$ is valid for $kx>0$ and $x>-2$.

Now, we simplify the given expression: $\log(kx)=2\log(x+2)=\log((x+2)^2) \implies kx=(x+2)^2$. The last equation is derived by taking away the outside logs from the previous equation.

Because $(x+2)^2$ is always non-negative, $kx$ must also be non-negative; therefore this takes care of the $kx>0$ condition as long as $k\neq0$, i.e. $k$ cannot be $0$. Now, we graph both $(x+2)^2$ (the green graph) and $kx$ (the red graph for $k=-1,k=-2,k=-3,k=8$) for $x>-2$. It is easy to see that all negative values of $k$ make the equation $\log(kx)=2\log(x+2)$ have only one solution.

However, there is also one positive value of $k$ that makes the equation only have one solution, as shown by the steepest line in the diagram. We can show that the slope of this line is a positive integer by setting the discriminant of the equation $(x+2)^2=kx$ to be $0$ and solving for $k$, which yields $k=8$.

When $1\leq k \leq 7$, $y=kx$ will intersect $y(x+2)^2$ twice, and when $k \geq 9$, $y=kx$ will never intersect $y(x+2)^2$.

Therefore, there are $500$ negative solutions and $1$ positive solution, for a total of $\boxed{501}$.

Solution 2

We use an algebraic approach. Since $\log(kx)=2\log(x+2)$, then $kx = (x+2)^2$ (the converse isn't necessarily true!), or $x^2+(4-k)x+4=0$. Our original equation has exactly one solution if and only if there is only one solution to the above equation, or one of the solutions is extraneous; it involves the computation of the log of a nonpositive number.

For the first case, this can only occur when it is a perfect square trinomial, or $k = 0, 8$. However, $k = 0$ results in $\log(0)$ on the left side, which is invalid. $k = 8$ yields $x = 2$, so that is one solution.

For the second case, we can use the quadratic formula. We have

$$x = \frac{k-4 \pm \sqrt{k^2-8k}}2,$$ so for there to be at least one real solution, the discriminant must be nonnegative, or $k < 0$ or $k > 8$.

Note that if $k > 8$, then both solutions to $x$ will be positive, and therefore both valid, which means all such $k$ are unsatisfactory.

We now wish to show that if $k < 0$, then exactly one solution works. Note that whenever $k < 0$, both "solutions" in $x$ are negative. One of the solutions to the equation is $x = \frac{k-4 + \sqrt{k^2-8k}}2$. We wish to prove that $x + 2 > 0$, or $x > -2$ (therefore the RHS in the original equation will be defined). Substituting, we have $\frac{k-4 + \sqrt{k^2-8k}}2 > -2$, or $\sqrt{k^2 - 8k} > -k$. Since both sides are positive, we can square both sides (if $k < 0$, then $-k > 0$) to get $k^2-8k > k^2$, or $8k < 0 \implies k < 0$, which was our original assumption, so this solution satisfies the original equation. The other case is when $x = \frac{k-4 - \sqrt{k^2-8k}}2$, which we wish to show is less that $-2$, or $\frac{k-4 - \sqrt{k^2-8k}}2 < -2 \implies k < \sqrt{k^2-8k}$. However, since the square root is defined to be positive, then this is always true, which implies that whenever $k < 0$, there is exactly one real solution that satisfies the original equation. Combining this with $k \in [-500, 500]$, we find that the answer is $500 + 1 = \boxed{501}$.

Note

The key to this solution is understanding that $\log(x+2)$ has a domain of $(-2, \infty),$ so in the second case, when there are two possible solutions of $x$ to $k<0,$ we notice that only the interval of the greater solution $(0,-2)$ works, which means that for any $k<0$ will have exactly one solution.

~mathboy282

Reworded Solution 2

Immediately we notice if $k$ is non-zero we must have $kx, (x+2) > 0$ for our sole solution $x = x_0$. Simplifying the logarithmic equation we get $kx = (x+2)^2 \rightarrow 0 = x^2 + (4-k)x + 4 \rightarrow x = \frac{k-4 \pm \sqrt{k^2 - 8k}}{2}$. Then $k \leq 0$ or $k \geq 8$. When $k = 8$ we have exactly one real solution (easily verifiable). Notice when $k > 8$, both solutions to $x$ are positive, and so all such $k$ is not satisfactory. When $k < 0$ it can be shown that the greater solution to $x$ is in the interval $(0,-2)$ and the lesser solution is in the interval $(-2,-\infty)$ which is satisfactory. Then $k = 8,-1,-2,...,-500 \rightarrow \boxed{501}$ satisfactory integer values of $k$.

~FRIDAY