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

(Created page with "<math>\textbf{Problem 7}</math> Find the number of integer values of <math>k</math> in the closed interval <math>[-500,500]</math> for which the equation <math>\log(kx)=2\log(...")
 
(→‎Solution 2)
 
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<math>\textbf{Problem 7}</math>
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==Problem==
 
Find the number of integer values of <math>k</math> in the closed interval <math>[-500,500]</math> for which the equation <math>\log(kx)=2\log(x+2)</math> has exactly one real solution.
 
Find the number of integer values of <math>k</math> in the closed interval <math>[-500,500]</math> for which the equation <math>\log(kx)=2\log(x+2)</math> has exactly one real solution.
  
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<math>\textbf{Problem 7 Solution}</math>
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==Solution 1==
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<asy>
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Label f;
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f.p=fontsize(5);
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xaxis(-3,3,Ticks(f,1.0));
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yaxis(-3,26,Ticks(f,1.0));
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real f(real x){return (x+2)^2;}
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real g(real x){return x*-1;}
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real h(real x){return x*-2;}
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real i(real x){return x*-3;}
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real j(real x){return x*8;}
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draw(graph(f,-2,3),green);
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draw(graph(g,-2,2),red);
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draw(graph(h,-2,1),red);
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draw(graph(i,-2,1/3),red);
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draw(graph(j,-0.25,3),red);
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</asy>
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Note the equation <math>\log(kx)=2\log(x+2)</math> is valid for <math>kx>0</math> and <math>x>-2</math>. <math>\log(kx)=2\log(x+2)=\log((x+2)^2)</math>. The equation <math>kx=(x+2)^2</math> is derived by taking away the outside logs from the previous equation. Because <math>(x+2)^2</math> is always non-negative, <math>kx</math> must also be non-negative; therefore this takes care of the <math>kx>0</math> condition as long as <math>k\neq0</math>, i.e. <math>k</math> cannot be <math>0</math>. Now, we graph both <math>(x+2)^2</math> (the green graph) and <math>kx</math> (the red graph for <math>k=-1,k=-2,k=-3,k=8</math>) for <math>x>-2</math>. It is easy to see that all negative values of <math>k</math> make the equation <math>\log(kx)=2\log(x+2)</math> have only one solution. However, there is also one positive value of <math>k</math> 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 <math>(x+2)^2=kx</math> to be <math>0</math> and solving for <math>k</math>. Therefore, there are <math>500</math> negative solutions and <math>1</math> positive solution, for a total of <math>\boxed{501}</math>.
 +
 
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==Solution 2==
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We use an algebraic approach. Since <math>\log(kx)=2\log(x+2)</math>, then <math>kx = (x+2)^2</math> (the converse isn't necessarily true!), or <math>x^2+(4-k)x+4=0</math>. 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 <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>.
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 +
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.
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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>.
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===Note===
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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
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===Reworded Solution 2===
 +
 
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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}}

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

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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