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this problem is a checkpoint for rewriting and simplifying expressions …

Question

this problem is a checkpoint for rewriting and simplifying expressions with integral and rational exponents. it will be referred to as checkpoint 3a. homework help
for parts (a) through (d), rewrite each expression. for parts (e) through (h), simplify each expression.
a. $sqrt5{x}$
b. $\frac{1}{x^{3}}$
c. $x^{2/3}$
d. $\frac{1}{sqrt{x}}$
e. $x^{-1}y^{-8}$
f. $(m^{2})^{-3/2}$
g. $(x^{3}y^{6})^{1/2}$
h. $(9x^{3}y^{6})^{-2}$

Explanation:

Step1: Use radical - exponent rule

For part a, $\sqrt[5]{x}=x^{\frac{1}{5}}$ according to the rule $\sqrt[n]{a}=a^{\frac{1}{n}}$.

Step2: Use negative - exponent rule

For part b, $\frac{1}{x^{3}} = x^{- 3}$ since $\frac{1}{a^{n}}=a^{-n}$.

Step3: Use radical - exponent rule

For part c, $x^{\frac{2}{3}}=\sqrt[3]{x^{2}}$ by the rule $a^{\frac{m}{n}}=\sqrt[n]{a^{m}}$.

Step4: Use radical - exponent rule

For part d, $\frac{1}{\sqrt{x}}=\frac{1}{x^{\frac{1}{2}}}=x^{-\frac{1}{2}}$ using $\sqrt{a}=a^{\frac{1}{2}}$ and $\frac{1}{a^{n}}=a^{-n}$.

Step5: Use negative - exponent rule

For part e, $x^{-1}y^{-8}=\frac{1}{xy^{8}}$ as $a^{-n}=\frac{1}{a^{n}}$.

Step6: Use power - of - a - power rule

For part f, $(m^{2})^{-\frac{3}{2}}=m^{2\times(-\frac{3}{2})}=m^{-3}=\frac{1}{m^{3}}$ using $(a^{m})^{n}=a^{mn}$ and $a^{-n}=\frac{1}{a^{n}}$.

Step7: Use power - of - a - product rule

For part g, $(x^{3}y^{6})^{\frac{1}{2}}=x^{3\times\frac{1}{2}}y^{6\times\frac{1}{2}}=x^{\frac{3}{2}}y^{3}=\sqrt{x^{3}}y^{3}$ using $(ab)^{n}=a^{n}b^{n}$ and $a^{\frac{m}{n}}=\sqrt[n]{a^{m}}$.

Step8: Use power - of - a - product rule

For part h, $(9x^{3}y^{6})^{-2}=9^{-2}x^{3\times(-2)}y^{6\times(-2)}=\frac{1}{81x^{6}y^{12}}$ using $(ab)^{n}=a^{n}b^{n}$, $a^{-n}=\frac{1}{a^{n}}$, and $9^{-2}=\frac{1}{9^{2}}=\frac{1}{81}$.

Answer:

a. $x^{\frac{1}{5}}$
b. $x^{-3}$
c. $\sqrt[3]{x^{2}}$
d. $x^{-\frac{1}{2}}$
e. $\frac{1}{xy^{8}}$
f. $\frac{1}{m^{3}}$
g. $x^{\frac{3}{2}}y^{3}$
h. $\frac{1}{81x^{6}y^{12}}$