Can k be equal to 4 in the equation k + 2 = 3^(n)?

  • MHB
  • Thread starter mathdad
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In summary, we are given the equation k + 2 = 3^(n), where n is a positive integer. We are asked to determine which of the given values for k (1, 4, 7, 25, 79) could not be a solution. After considering each value, we can conclude that k = 4 is not a possible solution, as there is no power of 3 that will yield 6 when added to 2. Therefore, the answer is B or 4.
  • #1
mathdad
1,283
1
If n is a positive integer and k + 2 = 3^(n), which of the following could NOT be a value of k?

A. 1
B. 4
C. 7
D. 25
E. 79

I say the answer is B or 4.

When k = 4, k + 2 becomes 6.
However, there is no power we can raised 3 to that will yield 6.

Is my reasoning correct?
 
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  • #2
RTCNTC said:
If n is a positive integer and k + 2 = 3^(n), which of the following could NOT be a value of k?

A. 1
B. 4
C. 7
D. 25
E. 79

I say the answer is B or 4.

When k = 4, k + 2 becomes 6.
However, there is no power we can raised 3 to that will yield 6.

Is my reasoning correct?
Looks good to me!

-Dan
 
  • #3
Or since 3^(n) is odd, then k can't be 4.
 

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