Question 9 (a) Write each of the following numbers in the form $3^k$, where $k \\in \\mathbb{Q}$ - Junior Cycle Mathematics - Question 9 - 2016

The number 1 can also be expressed in the form of powers of 3:

$1 = 3^0$

Here, $k = 0$.

To simplify $\sqrt{27}$, we find:

$\sqrt{27} = \sqrt{9 \cdot 3} = \sqrt{9} \cdot \sqrt{3} = 3 \sqrt{3}$

To express $3 \sqrt{3}$ in the form $3^k$, we can write:

$3 \sqrt{3} = 3^1 \cdot 3^{1/2} = 3^{1 + 1/2} = 3^{3/2}$

Thus, $k = \frac{3}{2}$.

To express $\frac{1}{\sqrt{3}}$ in the form $3^k$, we can rewrite:

$\frac{1}{\sqrt{3}} = 3^{-1/2}$

Therefore, $k = -\frac{1}{2}$.

To write $(-2n)^4$ in the form $a n^b$, we first expand:

$(-2n)^4 = (-2)^4 \cdot (n^4) = 16n^4$

Thus, we have $a = 16$ and $b = 4$.

An example of a value of $x$ is:

$x = -1$

In this case:

$\sqrt{x^2} = \sqrt{(-1)^2} = \sqrt{1} = 1$

So we have:

$x = -1 \\text{ and } \\sqrt{x^2} = 1$

These values are not equal.