Question

If $ {{n}^{K}}=64 $ and n and K are integers, then which of the following can not be a value of n?
A. 16
B. 8
C. 4
D. 2
E. -2

Answer 1

Hint: We will first write 64 in terms of power of 2 and then we will check the options one by one and find the value of K for the given value of n. If the value of K is an integer, then the option will be correct, otherwise it will be incorrect.

Complete step-by-step answer:
We have been given that $ {{n}^{K}}=64 $ , where n and K are integers. We have been asked to find the value that cannot be the value of n. We can write $ {{n}^{K}}=64 $ as $ {{n}^{K}}={{2}^{6}} $ also. So, in order to find the value that cannot be value of n, we will check each of the options one by one, whether we get the value of K as an integer for the given value of n.
A. 16
We can write 16 in terms of prime factors as,
 $ n=16=2\times 2\times 2\times 2={{2}^{4}} $
So, if n = 16, we can write the given equation as,
 $ \begin{align}
  & {{\left( {{2}^{4}} \right)}^{K}}={{2}^{6}} \\
 & \Rightarrow {{2}^{4K}}={{2}^{6}} \\
 & \Rightarrow 4K=6 \\
 & \Rightarrow K=\dfrac{6}{4} \\
 & \Rightarrow K=\dfrac{3}{2} \\
\end{align} $
We can see that $ K=\dfrac{3}{2} $ is not an integer. So, we cannot take n = 16 here. Therefore, option A is correct.
B. 8
If we consider 8, then the factors of 8 would be,
 $ n=8=2\times 2\times 2={{2}^{3}} $
So, when n = 8, then the equation can be written as,
 $ \begin{align}
  & {{\left( {{2}^{3}} \right)}^{K}}={{2}^{6}} \\
 & \Rightarrow {{2}^{3K}}={{2}^{6}} \\
 & \Rightarrow 3K=6 \\
 & \Rightarrow K=\dfrac{6}{3} \\
 & \Rightarrow K=2 \\
\end{align} $
Here, we can see that K = 2, which is an integer. So, we can consider n = 8 as the value. Therefore, option B is not correct.
C. 4
The factors of 4, would be,
 $ n=4=2\times 2={{2}^{2}} $
So, the equation would be,
 $ \begin{align}
  & {{\left( {{2}^{2}} \right)}^{K}}={{2}^{6}} \\
 & \Rightarrow {{2}^{2K}}={{2}^{6}} \\
 & \Rightarrow 2K=6 \\
 & \Rightarrow K=\dfrac{6}{2} \\
 & \Rightarrow K=3 \\
\end{align} $
As we can see, K =3, which is an integer. So, we can take n = 4 as the value. Therefore, option C is not correct.
D. 2
If n = 2, then the equation would be,
 $ \begin{align}
  & {{\left( 2 \right)}^{K}}={{2}^{6}} \\
 & \Rightarrow K=6 \\
\end{align} $
As K = 6 is an integer, we can take n = 2. Therefore, option D is also not correct.
E. -2
If we take n = -2, then the equation would be,
 $ \begin{align}
  & {{\left( -2 \right)}^{K}}={{2}^{6}} \\
 & \Rightarrow {{\left( -1 \right)}^{K}}{{2}^{K}}={{2}^{6}} \\
\end{align} $
Now, if K = 6, then, $ {{\left( -1 \right)}^{K}}=1 $ . So, it also satisfies the equation and the K = 6 is an integer, so -2 can also be taken as the value of n. Therefore, option E is also not correct.
Therefore, the correct answer for this question is option A.

Note: The first mistake that the students do in this question is to choose the options B, C, D and E as the correct answer to the question. But we should keep in mind that we have been asked to find, which of the given cannot be the value of n. So, we should be careful in reading the question and choosing the correct option.