Question

Can someone PLZZZZ HELP!!!!

Answer 1

Answer:

The solution is:

Part A. $\sqrt{5}^{\frac{7k}{3}})$ which is sqrt(5)^7k/3[/tex]

Part B. k = 18/7

Step-by-step explanation:

Part A.

To solve this part, we're going two use THREE important properties of exponents:

1. $(x^{n})^{m} = x^{nm}$

2. $\frac{x^{n}}{x^{m}} = x^{n-m}$

3. $\sqrt[n]{x^{m}} = x^{\frac{m}{n} }$

Let's work the numerator using the properties 1, 2 and 3:

$(\sqrt{5}^{3} )^{\frac{k}{9} } } = (\sqrt{5}^{3\frac{k}{9}}) = (\sqrt{5}^{\frac{k}{3}})$

Let's work the denominator using the properties 1, 2 and 3:

$(\sqrt{5}^{6} )^{-\frac{k}{3} } } = (\sqrt{5}^{6\frac{k}{3}}) = (\sqrt{5}^(2k))$

Now dividing the numerator by the denominator:

$\sqrt{5}^{\frac{k}{3}-(-2k)})=\sqrt{5}^{\frac{7k}{3}})$

Part B

if $5^{\frac{3}{2} } 5^{\frac{3}{2}} = \sqrt{5}^{\frac{7k}{3}})$

Then:

$5^{3} = 5^{\frac{7k}{6}})$

So $\frac{7k}{6}} = 3$

Solving for k, we have:

k = 18/7