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805+ (Hard)|   Exponents|      
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Russ19
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If n ≠ −2, for how many integer values of n is the equation (n + 2)^ 30 Jan 2021, 19:15
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D
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Difficulty:

95% (hard)

Question Stats:

30% (01:56) correct 70% (01:32) wrong based on 115 sessions

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If \( n ≠ −2\), for how many integer values of n is the equation \((n + 2)^{n^2}\)= \((n + 2)^{2n}\) true?

(A) none

(B) one

(C) two

(D) three

(E) infinitely many
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D
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If n ≠ −2, for how many integer values of n is the equation (n + 2)^ 30 Jan 2021, 20:30
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sjuniv32
If \( n ≠ −2\), for how many integer values of n is the equation \((n + 2)^{n^2}\)= \((n + 2)^{2n}\) true?

(A) none

(B) one

(C) two

(D) three

(E) infinitely many
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This will be true in two cases :
1) When the power/exponent is the same, as base is the same => \(n^2=2n\)
\(n^2-2n=0........n(n-2)=0\)
So, n can be 0 or 2.
2) The powers can be different when the base is 1.
So n+2=1....n=-1

Thus, three values possible.

D
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If n ≠ −2, for how many integer values of n is the equation (n + 2)^ 30 Jan 2021, 22:47
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When n + 2 = 1, then the expression will be true.

=> n + 2 = 1 or n = -1

Comparing powers, \(n^2 = 2n \)

=> n(n -2) = 0

=> n = 0, 2

Answer D
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If n ≠ −2, for how many integer values of n is the equation (n + 2)^ 06 Feb 2021, 14:00
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BrentGMATPrepNow, can you please resolve the problem above? When you explain a problem step by step, it gets easier for all. Thanks in advance!
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If n ≠ −2, for how many integer values of n is the equation (n + 2)^ 07 Feb 2021, 07:48
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sjuniv32
If \( n ≠ −2\), for how many integer values of n is the equation \((n + 2)^{n^2}\)= \((n + 2)^{2n}\) true?

(A) none
(B) one
(C) two
(D) three
(E) infinitely many
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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 \((n + 2)^{n^2}\)= \((n + 2)^{2n}\), then as long as \((n+2) \neq 0\), \((n+2) \neq 1\), and \((n+2) \neq -1\), then we can conclude that \(n^2 = 2n\)
Take: \(n^2 = 2n\)
Subtract \(2n\) from both sides to get: \(n^2 - 2n\)
Factor to get: to get: \(n(n - 2)\), so either \(n = 0\) or \(n = 2\) (two solutions so far)

Now we need to consider the possibility that the base, \((n+2)\), equals \(0\), \(1\) or \(-1\)

The condition that \( n ≠ −2\), eliminates the possibility that \((n+2) = 0\). So the base can't be zero.

Can the base, \((n+2)\), equal \(1\)?
We can see that \((n+2) = 1\), when \(n = -1\), so let's see if \(n = -1\) is an actual solution to the original equation.
Replace an to get: \(((-1) + 2)^{(-1)^2}\)= \(((-1) + 2)^{2(-1)}\)
Simplify to get: \((1)^{1}\)= \((1)^{-2}\) WORKS!!
So, \(n = -1\) is another solution (bringing the total number of solutions to three)

Finally, we need to test the possibility that the base, \((n+2)\), equals \(-1\)
We can see that \((n+2) = -1\), when \(n = -3\), so let's see if \(n = -3\) is a solution to the original equation.
Replace an to get: \(((-3) + 2)^{(-3)^2}\)= \(((-3) + 2)^{2(-3)}\)
Simplify to get: \((-1)^{9}\)= \((-1)^{-6}\)
Evaluate to get: \(-1\)= \(1\)
DOESN'T WORK


So, the solutions are \(n = 0\), \(n = 2\) and \(n = -1\)

Answer: C
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If n ≠ −2, for how many integer values of n is the equation (n + 2)^ 06 Dec 2022, 03:49
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Russ19
If \( n ≠ −2\), for how many integer values of n is the equation \((n + 2)^{n^2}\)= \((n + 2)^{2n}\) true?

(A) none

(B) one

(C) two

(D) three

(E) infinitely many
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0, -1, 2 are three integer values of n that satisfy the equation.

D is Correct
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If n ≠ −2, for how many integer values of n is the equation (n + 2)^ 22 Jun 2024, 21:07
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