Question

1. $(6v)^{1.5}$
2. $m^{-\frac{1}{2}}$

Explanation:

Problem 17: Simplify \((6v)^{1.5}\)

Step 1: Convert the decimal exponent to a fraction

We know that \(1.5=\frac{3}{2}\), so we can rewrite the expression as \((6v)^{\frac{3}{2}}\).

Step 2: Use the power - of - a - product rule \((ab)^n=a^n\times b^n\)

According to the rule \((ab)^n = a^n\times b^n\), where \(a = 6\), \(b=v\) and \(n=\frac{3}{2}\), we get \(6^{\frac{3}{2}}\times v^{\frac{3}{2}}\).

Step 3: Simplify \(6^{\frac{3}{2}}\)

We know that \(a^{\frac{m}{n}}=\sqrt[n]{a^m}\), so \(6^{\frac{3}{2}}=\sqrt{6^{3}}=\sqrt{216} = 6\sqrt{6}\) (or we can also calculate it as \((\sqrt{6})^{3}=6\sqrt{6}\)). And \(v^{\frac{3}{2}}=\sqrt{v^{3}}=v\sqrt{v}\) (or \(v^{\frac{3}{2}}=(v^{\frac{1}{2}})^{3}\)).
So \((6v)^{1.5}=6^{\frac{3}{2}}v^{\frac{3}{2}} = 6\sqrt{6}\times v\sqrt{v}=6v\sqrt{6v}\) (we can also combine the square - root terms: \(\sqrt{6^{3}v^{3}}=\sqrt{216v^{3}} = 6v\sqrt{6v}\))

Problem 18: Simplify \(m^{-\frac{1}{2}}\)

Step 1: Use the negative exponent rule \(a^{-n}=\frac{1}{a^{n}}\) (\(a

eq0\))
According to the negative exponent rule \(a^{-n}=\frac{1}{a^{n}}\), where \(a = m\) and \(n=\frac{1}{2}\), we can rewrite the expression as \(\frac{1}{m^{\frac{1}{2}}}\).

Step 2: Simplify \(m^{\frac{1}{2}}\)

We know that \(a^{\frac{1}{n}}=\sqrt[n]{a}\), so \(m^{\frac{1}{2}}=\sqrt{m}\). Then \(\frac{1}{m^{\frac{1}{2}}}=\frac{1}{\sqrt{m}}\). If we rationalize the denominator, we multiply the numerator and denominator by \(\sqrt{m}\), and we get \(\frac{\sqrt{m}}{m}\) (since \(\frac{1\times\sqrt{m}}{\sqrt{m}\times\sqrt{m}}=\frac{\sqrt{m}}{m}\) for \(m>0\)).

Answer:

s:

1. \((6v)^{1.5}=6v\sqrt{6v}\) (or \(6^{\frac{3}{2}}v^{\frac{3}{2}}\))
2. \(m^{-\frac{1}{2}}=\frac{1}{\sqrt{m}}\) (or \(\frac{\sqrt{m}}{m}\) after rationalizing the denominator)