Twist your brain with this challenging GRE problem! Then, leave a comment with your work and best guess. Good luck!

x^n + x^n + x^n = x^(n + 1)

Which of the following must be true?

I. n = 2

II. x = 3

III. n^x < x^n

(A) I only

(B) II and III

(C) I and III

(D) II only

(E) None of the above

See you on Friday for the answer and explanation. ðŸ™‚

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X=3 and n has infinite solution!

D is the right option since it satisfies the given equation and all the other option don’t so rest of the options are wrong.

Yes, (D) is the answer ðŸ™‚

Answer is D .. the option satisties the given equation and all the other options does not.

x^n+x^n+x^n = x^(n+1)

Solving this,

3 * x^n = x^n * x

=> x = 3.

Notice that we can cancel x^n term, which means for any value of n, the equation will hold. So answer has to be (D). We see something similar when two linear equations have infinite solutions.

Answer should be D. For X=3, for all values of “n” equation balances out

x can be both zero or 3.

so none of the above is must.

answer would be E.

E) none of the above

x cannot be 0

If x=0 and n=0, the left side will equal 3, but the right side equals 0. This fail to satisfy the equation

@Ahh Le

x and n can not be equal to 0 simultaneously because that would form a indeterminate form

that is “zero raised to power zero”.

which is not defined in world of mathematics

this question had some typos so it was confusing ðŸ™‚

Rearranging, we get

x^(n+1)-3x^n=0

x^n(x-3)=0

For this to be true, one of the following conditions must be true:

–> x=0, n can be any real number

–> x=3

However, the questions states that which of the following must be true, so while any of the options given can be true for the given statement to be valid, there is a possibility that the other condition is what is actually making the statement valid.

So there is no surety about which condition must be true.

Hence, the answer is (E).

the right answer is B both 2 and 3 is right ….

using the trying number technique by trying both the even and odd possibility

the first possibility 1* N cannot equal 2 as when u try 4 as a base for the square 2 both sides of the equation will not be equal .

on the other hand using the same technique of trying number u will find always that 2 and 3 must be true

III. Cannot be right.

The answer is (D). Not (B).

Here is why…

If x=3 and n can equal any number,

3^3+3^3+3^3=3(3+1)

27+27+27=3^4

81=81

Now for part III.

n^x < x^n

3^3 < 3^3

9 < 9 (This is not right, 9 cannot be less than 9)

Therefore…the correct answer is (D.)

The answer is II (x = 3).

x^n + x^n + x^n = x^(n + 1)

=> 3 * x^n = x * x^n

=> x = 3 (x^n at both sides of the equation is cancelled out)

Answer is (D) : II only

x^n + x^n + x^n = x^(n+1)

=> 3.x^n = (x^n).(x)

=>3 = x

also, 3^n and n^3 will be equal for n=3; therefore III is wrong

the eqn can be written as 3x^n =x^(n+1)

x=3 satisfies he eqn for any power for ex: x=10 gives 3(3^10)=3^11–>3^11=3^11 same for negative and fractional powers

n=2 need not be true for all values of x, but n=2 goes well for x=3..hence n=2 is not correct

3rd condition works for x=3. but not for different values of x and n

hence d is the answer

x^n + x^n + x^n = x^(n + 1)

3*(x^n) = x^(n+1)

=>

x=3

n=real number

for n=3, III is incorrect

D option

I would go with D. n =2 doesn’t guarantee a 0 on the RHS of the equation. Similarly, one cannot draw a determinate conclusion about n and x in the third option. But, x=3 guarantees an RHS=0. Hence the guess

Answer is “D”

(E) none of the above.

Since I and II are true.