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

(fixed \LaTeX)
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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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==Solution==
+
==Solution 1==
 +
<math>\textbf{Problem 7 Solution}</math>
 +
<asy>
 +
Label f;
 +
f.p=fontsize(5);
 +
xaxis(-3,3,Ticks(f,1.0));
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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;}
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real j(real x){return x*8;}
 +
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>
 +
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>.
 +
 
 +
==Solution 2==
 
<math>kx=(x+2)^2</math>
 
<math>kx=(x+2)^2</math>
 
<math>x^2+(4-k)x+4=0 ...(1)</math>
 
<math>x^2+(4-k)x+4=0 ...(1)</math>

Revision as of 16:11, 23 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

$\textbf{Problem 7 Solution}$ [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

$kx=(x+2)^2$ $x^2+(4-k)x+4=0 ...(1)$

the equation has solution so

$D=(4-k)^2-16=k(k-8)\geq0$

so $k<0$ or $k\geq8$ because k can't be zero or the original equation will be meaningless. there are 3 cases

1:$k=8$ then $x=2$, which is satisified the question.

2:$k<0$ then one solution of the equation(1) should be in $(-2,0)$ and another is out of it or the origin equation will be meanless. then we get 2 inequalities $-2<\frac{k-4+\sqrt{k(k-8)}}{2}<0$

$\frac{k-4-\sqrt{k(k-8)}}{2}<-2$ notice $k<0<\sqrt{k(k-8)}$ and $(4-k)^2=k(k-8)+16>k(k-8)$ we know in this case, there is always and only one solution for the orign equation.

3:$k>8$ similar to case2 we can get inequality $\frac{k-4-\sqrt{k(k-8)}}{2}<0<\frac{k-4+\sqrt{k(k-8)}}{2}$ and there are always 2 solution for the origin equation, so this case is not satisfied.

so we get $k<0$ or $k=8$

because k belong to $[-500,500]$, the answer is $\boxed{501}$

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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