Question

rewrite each expression below so that there are no negative exponents or parentheses remaining.

a. \\(\frac{4x^{18}}{(2x^{22})^0}\\)

b. \\((s^4tu^2)(s^7t^{-1})\\)

c. \\((3w^{-2})^4\\)

d. \\(m^{-3}\\)

Explanation:

Step1: Solve part a

Recall that any non - zero number to the power of 0 is 1, i.e., \(a^{0}=1\) for \(a
eq0\). Here, \((2x^{22})^{0} = 1\) (since \(2x^{22}
eq0\) as long as \(x\) is a real number and \(x
eq0\), but even if \(x = 0\), the original expression \(\frac{4x^{18}}{(2x^{22})^{0}}\) would be undefined as we can't divide by 0 when \(x = 0\) and \((2x^{22})^{0}=1\) when \(x
eq0\)). So \(\frac{4x^{18}}{(2x^{22})^{0}}=\frac{4x^{18}}{1}=4x^{18}\).

Step2: Solve part b

When multiplying two terms with the same base, we add the exponents. For the base \(s\), we have \(s^{4}\times s^{7}=s^{4 + 7}=s^{11}\). For the base \(t\), we have \(t^{1}\times t^{-1}=t^{1+( - 1)}=t^{0}=1\) (since \(a^{m}\times a^{n}=a^{m + n}\) and \(a^{0}=1\) for \(a
eq0\)). For the base \(u\), we have \(u^{2}\). So \((s^{4}tu^{2})(s^{7}t^{-1})=s^{11}\times1\times u^{2}=s^{11}u^{2}\).

Step3: Solve part c

Use the power of a product rule \((ab)^{n}=a^{n}b^{n}\) and the power of a power rule \((a^{m})^{n}=a^{mn}\). First, \((3w^{-2})^{4}=3^{4}\times(w^{-2})^{4}\). Calculate \(3^{4}=81\) and \((w^{-2})^{4}=w^{-2\times4}=w^{-8}\). Then, use the rule \(a^{-n}=\frac{1}{a^{n}}\) to rewrite \(w^{-8}\) as \(\frac{1}{w^{8}}\). So \((3w^{-2})^{4}=\frac{81}{w^{8}}\).

Step4: Solve part d

Use the rule \(a^{-n}=\frac{1}{a^{n}}\). So \(m^{-3}=\frac{1}{m^{3}}\).

Answer:

a. \(4x^{18}\)

b. \(s^{11}u^{2}\)

c. \(\frac{81}{w^{8}}\)

d. \(\frac{1}{m^{3}}\)