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What is sum of all solutions to the equation: [m]x^{(x^2 + 2)} =
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12 Jan 2021, 06:34
12 Jan 2021, 06:34
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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
Re: What is sum of all solutions to the equation: [m]x^{(x^2 + 2)} =
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12 Jan 2021, 09:13
12 Jan 2021, 09:13
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There are two possibilities straight up;
1) x=1 as all exponents of 1 are 1
or
2) the powers on both sides of the equation are equal
ie. x^2 + 2 = 4x + 14
upon solving,
gives (x-6)(x+2)=0
ie. x= 6, -2
therefore the sum total of the values of x is 6+(-2)+1=5
1) x=1 as all exponents of 1 are 1
or
2) the powers on both sides of the equation are equal
ie. x^2 + 2 = 4x + 14
upon solving,
gives (x-6)(x+2)=0
ie. x= 6, -2
therefore the sum total of the values of x is 6+(-2)+1=5
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: What is sum of all solutions to the equation: [m]x^{(x^2 + 2)} =
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12 Jan 2021, 10:52
12 Jan 2021, 10:52
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GreenlightTestPrep wrote:
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
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
