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

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In the equation k + 2 = 3^(n), where n is a positive integer, k cannot equal 4 because it results in k + 2 equaling 6, which is not a power of 3. The discussion highlights that since 3 raised to any integer power is odd, k must also be odd to maintain the equation's integrity. Therefore, options like 1, 7, 25, and 79 are valid values for k, while 4 is not. The reasoning presented confirms that k = 4 is indeed an invalid solution. The conclusion emphasizes the importance of the parity of k in relation to the powers of 3.
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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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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
 
Or since 3^(n) is odd, then k can't be 4.
 
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