2017 AIME II Problems/Problem 7
Problem
Find the number of integer values of in the closed interval for which the equation has exactly one real solution.
Solution 1
Note the equation is valid for and . . The equation is derived by taking away the outside logs from the previous equation. Because is always non-negative, must also be non-negative; therefore this takes care of the condition as long as , i.e. cannot be . Now, we graph both (the green graph) and (the red graph for ) for . It is easy to see that all negative values of make the equation have only one solution. However, there is also one positive value of 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 to be and solving for . Therefore, there are negative solutions and positive solution, for a total of .
Solution 2
We use an algebraic approach. Since , then (the converse isn't necessarily true!), or . 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 . However, results in on the LHS, which is invalid. yields , so that is one solution.
For the second case, we can use the quadratic formula. We have so in order for there to be at least one real solution, the discriminant must be nonnegative, or or .
Note that if , then both solutions to will be positive, and therefore both valid, which means all such are unsatisfactory.
We now wish to show that if , then there is exactly one solution that works. Note that whenever , both "solutions" in are negative. One of the solutions to the equation is . We wish to prove that , or (therefore the RHS in the original equation will be defined). Substituting, we have , or . Since both sides are positive, we can square both sides (if , then ) to get , or , which was our original assumption, so this solution satisfies the original equation. The other case is when , which we wish to show is less that , or . However, since the square root is defined to be positive, then this is always true, which implies that whenever , there is exactly one real solution that satisfies the original equation. Combining this with , we find that the answer is .
Note
The key to this solution is understanding that has a domain of so in the second case, when there are two possible solutions of to we notice that only the interval of the greater solution works, which means that for any will have exactly one solution.
~mathboy282
Reworded Solution 2
Immediately we notice is non-zero, in fact we must have for our sole solution . Simplifying the logarithmic equation we get . Then or . When we have exactly one real solution (easily verifiable). Notice when , both solutions to are positive, and so all such are not satisfactory. When it can be shown that the greater solution to is in the interval and the lesser solution is in the interval which is satisfactory. Then satisfactory integer values of .
~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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