Question

consider harpers steps to write an equivalent expression. harper made an error in one of her steps. which step contains harpers error? step 1: 8^{\frac{1}{2}}cdot64^{\frac{1}{2}} step 2: 8^{\frac{1}{2}}cdot(8^{2})^{\frac{1}{4}} step 3: 8^{\frac{1}{2}}cdot8^{\frac{2}{4}} step 4: 8^{\frac{3}{6}} final: 8^{\frac{1}{2}} what is the correct final answer?

Explanation:

Step1: Analyze the original expression $\sqrt{8}\cdot64^{\frac{1}{4}}$

We know that $\sqrt{8}=8^{\frac{1}{2}}$ and $64 = 8^{2}$, so $64^{\frac{1}{4}}=(8^{2})^{\frac{1}{4}}$.

Step2: Simplify $(8^{2})^{\frac{1}{4}}$

Using the power - of - a - power rule $(a^{m})^{n}=a^{mn}$, we have $(8^{2})^{\frac{1}{4}}=8^{\frac{2}{4}} = 8^{\frac{1}{2}}$. Then the original expression becomes $8^{\frac{1}{2}}\cdot8^{\frac{1}{2}}$.

Step3: Simplify $8^{\frac{1}{2}}\cdot8^{\frac{1}{2}}$

Using the product rule $a^{m}\cdot a^{n}=a^{m + n}$, we get $8^{\frac{1}{2}+\frac{1}{2}}=8^{1}=8$.

Step4: Check for errors

The error is in Step 4. When multiplying $8^{\frac{1}{2}}\cdot8^{\frac{1}{2}}$, we should get $8^{\frac{1 + 1}{2}}=8^{1}$ not $8^{\frac{3}{6}}$.

Answer:

Which step contains Harper's error? Step 4
What is the correct final answer? 8