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Use quotient property and negative Exponent Property to generate an equivalent expression for \sqrt((49)/(7^(-3))). What will the rational exponent be for the new expression?

7^

3 answers

To simplify the expression \(\sqrt{\frac{49}{7^{-3}}}\), we can use the quotient property and the negative exponent property.

Step 1: Rewriting the expression with properties.

The expression can be rewritten as:

\[ \sqrt{\frac{49}{7^{-3}}} = \sqrt{49} \div \sqrt{7^{-3}} \]

Step 2: Simplifying \(\sqrt{49}\).

We know that:

\[ \sqrt{49} = 7 \]

Step 3: Simplifying \(\sqrt{7^{-3}}\).

Using the property of exponents that states \(\sqrt{a^b} = a^{\frac{b}{2}}\), we have:

\[ \sqrt{7^{-3}} = 7^{-\frac{3}{2}} \]

Step 4: Putting it all together.

Now we can represent \(\sqrt{49}\) over \(\sqrt{7^{-3}}\):

\[ \sqrt{\frac{49}{7^{-3}}} = \frac{7}{7^{-\frac{3}{2}}} \]

Step 5: Applying the Quotient Property of Exponents.

Using the quotient property \(a^m \div a^n = a^{m-n}\):

\[ \frac{7}{7^{-\frac{3}{2}}} = 7^{1 - (-\frac{3}{2})} = 7^{1 + \frac{3}{2}} = 7^{\frac{2}{2} + \frac{3}{2}} = 7^{\frac{5}{2}} \]

So, the simplified expression is:

\[ 7^{\frac{5}{2}} \]

Conclusion: The rational exponent for the new expression is \(\frac{5}{2}\).

wrong

Let's go through the simplification of \(\sqrt{\frac{49}{7^{-3}}}\) step by step more carefully.

Step 1: Rewrite the expression

The expression \(\sqrt{\frac{49}{7^{-3}}}\) can be rewritten using the property of square roots:

\[ \sqrt{\frac{49}{7^{-3}}} = \frac{\sqrt{49}}{\sqrt{7^{-3}}} \]

Step 2: Simplify \(\sqrt{49}\)

We know that:

\[ \sqrt{49} = 7 \]

Step 3: Simplify \(\sqrt{7^{-3}}\)

Now we simplify \(\sqrt{7^{-3}}\) using the property of exponents:

\[ \sqrt{7^{-3}} = 7^{-\frac{3}{2}} \]

Step 4: Divide the expressions

Now we can combine the results:

\[ \frac{\sqrt{49}}{\sqrt{7^{-3}}} = \frac{7}{7^{-\frac{3}{2}}} \]

Step 5: Apply the Quotient Property

Now, when we divide these two powers of 7, we use the quotient property \(a^m \div a^n = a^{m-n}\):

\[ \frac{7}{7^{-\frac{3}{2}}} = 7^{1 - (-\frac{3}{2})} = 7^{1 + \frac{3}{2}} \]

Step 6: Simplify the exponent

To simplify this further:

\[ 1 = \frac{2}{2} \quad \text{so} \quad 1 + \frac{3}{2} = \frac{2}{2} + \frac{3}{2} = \frac{5}{2} \]

Thus, we find:

\[ \frac{7}{7^{-\frac{3}{2}}} = 7^{\frac{5}{2}} \]

Conclusion

The final equivalent expression for \(\sqrt{\frac{49}{7^{-3}}}\) is:

\[ 7^{\frac{5}{2}} \]

The rational exponent for the new expression is indeed \(\frac{5}{2}\).

If you were looking for something else or if there was a misunderstanding regarding what was considered "wrong", please clarify and I'll be happy to assist further!