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Difference between revisions of "2017 AIME II Problems/Problem 7"

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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 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>. Note that if <math>k > 8</math>, then both solutions will be positive, and therefore both valid. Therefore, <math>k < 0</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 \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>.
 
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>.
  

Revision as of 14:40, 24 March 2017

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 will be positive, and therefore both valid. Therefore, $k < 0$. 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 $500 + 1 = \boxed{501}$.

See Also

2017 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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