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$Express the following fraction in simplest form, only using positive exponents. ((4a^(-5)k^(-4))^(-5))/(2k^(-9)) Answer:$

Express the following fraction in simplest form, only using positive exponents. $\newline$ $\frac{\left(4 a^{-5} k^{-4}\right)^{-5}}{2 k^{-9}}$ $\newline$ Answer:

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Math Problems

Algebra 1

Multiplication with rational exponents

Full solution

Q. Express the following fraction in simplest form, only using positive exponents.

\newline

\frac{\left(4 a^{-5} k^{-4}\right)^{-5}}{2 k^{-9}}

\newline

Answer:

Write Expression, Identify Exponents: Write down the given expression and identify the negative exponents. $\newline$ The given expression is $\left(\frac{(4a^{-5}k^{-4})^{-5}}{2k^{-9}}\right)$ . $\newline$ We need to simplify this expression and convert all negative exponents to positive exponents.
Apply Negative Exponent Rule: Apply the negative exponent rule, which states that $a^{-n} = \frac{1}{a^n}$ , to the numerator. $\newline$ $((4a^{-5}k^{-4})^{-5})$ becomes $(\frac{1}{4a^5k^4})^5$ .
Apply Power Rule: Apply the power rule, which states that $(a/b)^n = a^n/b^n$ , to the numerator. $(1/(4a^5k^4))^5$ becomes $1^5/(4^5a^{(5*5)}k^{(4*5)})$ .
Simplify Numerator: Simplify the numerator using the power of $1$ and the power rule for exponents. $\newline$ $1^5/(4^5a^{5*5}k^{4*5})$ becomes $1/(1024a^{25}k^{20})$ .
Apply Negative Exponent Rule to Denominator: Apply the negative exponent rule to the denominator, which states that $k^{-n} = \frac{1}{k^n}$ . $2k^{-9}$ becomes $\frac{2}{1} \times \frac{1}{k^9}$ , which simplifies to $\frac{2}{k^9}$ .
Combine into Single Fraction: Combine the numerator and the denominator into a single fraction. $\newline$ $(\frac{1}{1024a^{25}k^{20}}) / (\frac{2}{k^{9}})$ becomes $(\frac{1}{1024a^{25}k^{20}}) \cdot (\frac{k^{9}}{2})$ .
Multiply Fractions: Multiply the fractions. $\newline$ $(\frac{1}{1024a^{25}k^{20}}) \times (\frac{k^{9}}{2})$ becomes $\frac{k^{9}}{2048a^{25}k^{20}}$ .
Simplify Fraction: Simplify the fraction by canceling out the common $k$ terms. $\frac{k^9}{2048a^{25}k^{20}}$ becomes $\frac{1}{2048a^{25}k^{(20-9)}}$ , which simplifies to $\frac{1}{2048a^{25}k^{11}}$ .