simplify: $\frac{w^{7/8}}{w^{1/8}}$ \na $w^{6}$ \nb $sqrt8{w^{5}}$ \nc $w^{7}$ \nd $sqrt4{w^{3}}$
simplify: $\frac{w^{7/8}}{w^{1/8}}$ \na $w^{6}$ \nb $sqrt8{w^{5}}$ \nc $w^{7}$ \nd $sqrt4{w^{3}}$
Answer
Explanation:
Step1: Use exponent - division rule
When dividing two terms with the same base (a^m\div a^n=a^{m - n}), here (a = w), (m=\frac{7}{8}), and (n=\frac{1}{8}). So (\frac{w^{7/8}}{w^{1/8}}=w^{\frac{7}{8}-\frac{1}{8}}).
Step2: Calculate the exponent
(\frac{7}{8}-\frac{1}{8}=\frac{7 - 1}{8}=\frac{6}{8}=\frac{3}{4}). So (w^{\frac{7}{8}-\frac{1}{8}}=w^{3/4}).
Step3: Convert to radical form
By the rule (a^{m/n}=\sqrt[n]{a^m}), when (a = w), (m = 3), and (n = 4), (w^{3/4}=\sqrt[4]{w^{3}}).
Answer:
D. (\sqrt[4]{w^{3}})