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Simplify. Assume all variables are positive. $\newline$ $\frac{v^{\frac{7}{3}}}{v^{\frac{8}{3}}}$ $\newline$ Write your answer in the form $A$ or $\frac{A}{B}$ , where $A$ and $B$ are constants or variable expressions that have no variables in common. All exponents in your answer should be positive. $\newline$ ______

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Simplify expressions involving rational exponents

Full solution

Q. Simplify. Assume all variables are positive.

\newline

\frac{v^{\frac{7}{3}}}{v^{\frac{8}{3}}}

\newline

Write your answer in the form

A

or

\frac{A}{B}

, where

A

and

B

are constants or variable expressions that have no variables in common. All exponents in your answer should be positive.

\newline

______

Identify operation: We have the expression: $\newline$ $v^{\frac{7}{3}}/v^{\frac{8}{3}}$ $\newline$ Which operation will be applied with the exponents? $\newline$ When dividing powers with the same base, the exponents are subtracted.
Apply rule for division: Apply the rule for dividing powers with the same base: $\newline$ $v^{\frac{7}{3}}/v^{\frac{8}{3}} = v^{\frac{7}{3} - \frac{8}{3}}$ $\newline$ Perform the subtraction of the exponents. $\newline$ $v^{-\frac{1}{3}}$
Rewrite using reciprocal: Since we want the exponent to be positive, we can rewrite the expression using the reciprocal of $v$ : $v^{(-1/3)} = 1/v^{(1/3)}$

More problems from Simplify expressions involving rational exponents

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h(x)=\log _{2}(x)

\newline

Can we use the mean value theorem to say the equation

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has a solution where

1<x<2

\newline

1

\newline

(A) No, since the function is not differentiable on that interval.

\newline

(B) No, since the average rate of change of

h

over the interval

1 \leq x \leq 2

1

\newline

(C) Yes, both conditions for using the mean value theorem have been met.

Posted 9 months ago

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Question

\begin{array}{l} g(x)=\sqrt{2 x-9} \\ g^{\prime}(x)=? \end{array}

\newline

1

\newline

\frac{\sqrt{2 x-9}}{2}

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\frac{1}{\sqrt{2 x-9}}

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\frac{1}{2 \sqrt{2 x-9}}

\newline

\frac{1}{\sqrt{x}}

Posted 9 months ago

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