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

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

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 $\boxed{501}$

2017 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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2017 AIME II Problems/Problem 7

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