Which expressions are equivalent to (k^((1)/(8)))^(-1) ? Choose all answers that apply: A (k^(-1))^((1)/(8)) B (root(8)(k))^(-1) C k^(-(1)/(8)) D None of the above

Understand properties of exponents: Understand the properties of exponents. When raising a power to another power, you multiply the exponents. So, (k^(1/8))^(-1) should be simplified by multiplying the exponents (1/8) and (-1).Apply exponent multiplication rule: Apply the exponent multiplication rule. (k^(1/8))^(-1) = k^((1/8)*(-1)) = k^(-1/8)Compare with given options: Compare the result with the given options. Option A: (k^(-1))^(1/8) is not equivalent because it suggests taking the reciprocal of k first and then the 8th root, which is not the same as k^(-1/8). Option B: (root(8)(k))^(-1) is equivalent because taking the 8th root of k and then taking the reciprocal is the same as raising k to the power of -1/8. Option C: k^(-1/8) is exactly what we found in Step 2, so it is equivalent. Option D: ＂None of the above＂ is not correct because we have found equivalent expressions in the options provided.

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Which expressions are equivalent to
(k^((1)/(8)))^(-1) ?
Choose all answers that apply:
A
(k^(-1))^((1)/(8))
B
(root(8)(k))^(-1)
C
k^(-(1)/(8))
D None of the above

Which expressions are equivalent to $\left(k^{\frac{1}{8}}\right)^{-1}$ ? $\newline$ Choose all answers that apply: $\newline$ A $\left(k^{-1}\right)^{\frac{1}{8}}$ $\newline$ B $(\sqrt[8]{k})^{-1}$ $\newline$ C $k^{-\frac{1}{8}}$ $\newline$ D None of the above

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Math Problems

Grade 7

Identify equivalent linear expressions I

Full solution

Q. Which expressions are equivalent to

\left(k^{\frac{1}{8}}\right)^{-1}

\newline

Choose all answers that apply:

\newline

\left(k^{-1}\right)^{\frac{1}{8}}

\newline

(\sqrt[8]{k})^{-1}

\newline

k^{-\frac{1}{8}}

\newline

D None of the above

Understand properties of exponents: Understand the properties of exponents. When raising a power to another power, you multiply the exponents. So, $(k^{1/8})^{-1}$ should be simplified by multiplying the exponents $(1/8)$ and $(-1)$ .
Apply exponent multiplication rule: Apply the exponent multiplication rule. $\newline$ $(k^{\frac{1}{8}})^{-1} = k^{(\frac{1}{8})\cdot(-1)} = k^{-\frac{1}{8}}$
Compare with given options: Compare the result with the given options. $\newline$ Option A: $(k^{-1})^{\frac{1}{8}}$ is not equivalent because it suggests taking the reciprocal of $k$ first and then the $8$ th root, which is not the same as $k^{-\frac{1}{8}}$ . $\newline$ Option B: $(\sqrt[8]{k})^{-1}$ is equivalent because taking the $8$ th root of $k$ and then taking the reciprocal is the same as raising $k$ to the power of $-\frac{1}{8}$ . $\newline$ Option C: $k^{-\frac{1}{8}}$ is exactly what we found in Step $k$ $0$ , so it is equivalent. $\newline$ Option D: ＂None of the above＂ is not correct because we have found equivalent expressions in the options provided.