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11 Jan 2021, 15:11
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
37% (01:41) correct
63%
(01:22)
wrong
based on 260
sessions
History
Date
Time
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Not Attempted Yet
What is sum of all solutions to the equation: \(x^{(x^2 + 2)} = x^{(4x + 14)}\) ?
(A) -4
(B) 4
(C) 5
(D) 6
(E) 8
(A) -4
(B) 4
(C) 5
(D) 6
(E) 8
12 Jan 2021, 07:23
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BrentGMATPrepNow
Key concept: If \(b^x = b^y\), then \(x = y \) (as long as \(b \neq 0\), \(b \neq 1\), and \(b \neq -1\))
The provisos here a very important.
For example, if \(0^x = 0^y\), we can't then conclude that \(x=y\)
Now on to the solution.....
If \(x^{(x^2 + 2)} = x^{(4x + 14)}\), then we can conclude that: \(x^2 + 2 = 4x + 14\)
Rewrite as follows: \(x^2 - 4x - 12= 0\)
Factor: \((x-6)(x+2)=0\)
So, two solutions are \(x=6 \) and \(x = -2\)
At this point, we need to test the "proviso" values (\(x=0\), \(x=1\) and \(x=-1\))
Test \(x=0\) to get: \(0^{0^2 + 2} = 0^{4(0) + 14}\). Works! So, another solution is \(x=0\)
Test \(x=1\) to get: \(1^{1^2 + 2} = 1^{4(1) + 14}\). Works! So, another solution is \(x=1\)
Test \(x=-1\) to get: \((-1)^{(-1)^2 + 2} = (-1)^{4(-1) + 14}\). DOESN'T work, So, \(x=-1\) is NOT a solution
So, the sum of the solutions \(= 6 + (-2) + 0 + 1 = 5\)
Answer: C
General Discussion
11 Jan 2021, 15:24
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base is same, so equate exponents and solve for x
x^2 + 2 = 4x + 14
x= 6, -2
sum = 4
Ans. B
x^2 + 2 = 4x + 14
x= 6, -2
sum = 4
Ans. B
