Latest revision as of 22:03, 5 January 2024

Problem

Find the number of integer values of $k$ in the closed interval $[-500,500]$ for which the equation $\log(kx)=2\log(x+2)$ has exactly one real solution.

Solution 1

$[asy] Label f; f.p=fontsize(5); xaxis(-3,3,Ticks(f,1.0)); yaxis(-3,26,Ticks(f,1.0)); real f(real x){return (x+2)^2;} real g(real x){return x*-1;} real h(real x){return x*-2;} real i(real x){return x*-3;} real j(real x){return x*8;} draw(graph(f,-2,3),green); draw(graph(g,-2,2),red); draw(graph(h,-2,1),red); draw(graph(i,-2,1/3),red); draw(graph(j,-0.25,3),red); [/asy]$ Note the equation $\log(kx)=2\log(x+2)$ is valid for $kx>0$ and $x>-2$ . $\log(kx)=2\log(x+2)=\log((x+2)^2)$ . The equation $kx=(x+2)^2$ 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 $k$ . 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, we note that this can only occur when it is a perfect square trinomal, or $k = 0, 8$ . However, $k = 0$ results in $\log(0)$ on the LHS, 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 in order 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 there is exactly one solution that 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 $k$ is non-zero, in fact 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 $k$ are 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

@@ Line 27: / Line 27: @@
 For the first case, we note that this can only occur when it is a perfect square trinomal, or <math>k = 0, 8</math>. However, <math>k = 0</math> results in <math>\log(0)</math> on the LHS, which is invalid. <math>k = 8</math> yields <math>x = 2</math>, so that is one solution.
-For the second case, we can use the quadratic formula. We have <cmath>x = \frac{k-4 \pm \sqrt{k^2-8k}}2</cmath>, so in order for there to be at least one real solution, the discriminant must be nonnegative, or <math>k < 0</math> or <math>k > 8</math>. Note that if <math>k > 8</math>, then both solutions will be positive, and therefore both valid. Therefore, <math>k < 0</math>.
+For the second case, we can use the quadratic formula. We have <cmath>x = \frac{k-4 \pm \sqrt{k^2-8k}}2,</cmath> so in order for there to be at least one real solution, the discriminant must be nonnegative, or <math>k < 0</math> or <math>k > 8</math>.
-We now wish to show that if <math>k < 0</math>, then there is exactly one solution that works. Note that whenever <math>k < 0</math>, both "solutions" in <math>x</math> are negative. One of the solutions to the equation is <math>x = \frac{k-4 + \sqrt{k^2-8k}}2</math>. We wish to prove that <math>x + 2 > 0</math>, or <math>x > -2</math> (therefore the RHS in the original equation will be defined).Substituting, we have <math>\frac{k-4 + \sqrt{k^2-8k}}2 > -2</math>, or <math>\sqrt{k^2 - 8k} > -k</math>. Since both sides are positive we can square both sides (if <math>k < 0</math>, then <math>-k > 0</math>) to get <math>k^2-8k > k^2</math>, or <math>8k < 0 \imples k < 0</math>, which was our original assumption, so this solution satisfies the original equation. The other case is when <math>x = \frac{k-4 - \sqrt{k^2-8k}}2</math>, which we wish to show is less that <math>-2</math>, or <math>\frac{k-4 - \sqrt{k^2-8k}}2 < -2 \implies k < \sqrt{k^2-8k}</math>. However, since the square root is defined to be positive, then this is always true, which implies that whenever <math>k < 0</math>, there is exactly one real solution that satisfies the original equation. Combining this with <math>k \in [-500, 500]</math>, we find that the answer is <math>500 + 1 = \boxed{501}</math>.
+Note that if <math>k > 8</math>, then both solutions to <math>x</math> will be positive, and therefore both valid, which means all such <math>k</math> are unsatisfactory.
+We now wish to show that if <math>k < 0</math>, then there is exactly one solution that works. Note that whenever <math>k < 0</math>, both "solutions" in <math>x</math> are negative. One of the solutions to the equation is <math>x = \frac{k-4 + \sqrt{k^2-8k}}2</math>. We wish to prove that <math>x + 2 > 0</math>, or <math>x > -2</math> (therefore the RHS in the original equation will be defined). Substituting, we have <math>\frac{k-4 + \sqrt{k^2-8k}}2 > -2</math>, or <math>\sqrt{k^2 - 8k} > -k</math>. Since both sides are positive, we can square both sides (if <math>k < 0</math>, then <math>-k > 0</math>) to get <math>k^2-8k > k^2</math>, or <math>8k < 0 \implies k < 0</math>, which was our original assumption, so this solution satisfies the original equation. The other case is when <math>x = \frac{k-4 - \sqrt{k^2-8k}}2</math>, which we wish to show is less that <math>-2</math>, or <math>\frac{k-4 - \sqrt{k^2-8k}}2 < -2 \implies k < \sqrt{k^2-8k}</math>. However, since the square root is defined to be positive, then this is always true, which implies that whenever <math>k < 0</math>, there is exactly one real solution that satisfies the original equation. Combining this with <math>k \in [-500, 500]</math>, we find that the answer is <math>500 + 1 = \boxed{501}</math>.
+===Note===
+The key to this solution is understanding that <math>\log(x+2)</math> has a domain of <math>(-2, \infty),</math> so in the second case, when there are two possible solutions of <math>x</math> to <math>k<0,</math> we notice that only the interval of the greater solution <math>(0,-2)</math> works, which means that for any <math>k<0</math> will have exactly one solution.
+~mathboy282
+===Reworded Solution 2===
+Immediately we notice <math>k</math> is non-zero, in fact we must have <math>kx, (x+2) > 0</math> for our sole solution <math>x = x_0</math>. Simplifying the logarithmic equation we get <math>kx = (x+2)^2 \rightarrow 0 = x^2 + (4-k)x + 4 \rightarrow x = \frac{k-4 \pm \sqrt{k^2 - 8k}}{2}</math>. Then <math>k \leq 0</math> or <math>k \geq 8</math>. When <math>k = 8</math> we have exactly one real solution (easily verifiable). Notice when <math>k > 8</math>, both solutions to <math>x</math> are positive, and so all such <math>k</math> are not satisfactory. When <math>k < 0</math> it can be shown that the greater solution to <math>x</math> is in the interval <math>(0,-2)</math> and the lesser solution is in the interval <math>(-2,-\infty)</math> which is satisfactory. Then <math>k = 8,-1,-2,...,-500 \rightarrow \boxed{501}</math> satisfactory integer values of <math>k</math>.
+~FRIDAY
 =See Also=
 {{AIME box|year=2017|n=II|num-b=6|num-a=8}}
 {{MAA Notice}}

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