Question

PLEASE HELP. I have no idea what to do

Answer 1

Answer:

What is it i can probably help

Step-by-step explanation:

Answer 2

$since \to : x \sqrt[y]{ {n}^{z} } = x( {n}^{ \frac{z}{y} } ) \\ then \: lets \: solve \: both \: values \: separately \to : \\ \underline{ \boxed{solution \: to \: \boxed{a}}} \\ 120 \sqrt[3]{ {n}^{a} } = 3 \sqrt{n} (40 \sqrt[6]{n} ) \\ 120 \sqrt[3]{ {n}^{a} } = 3 \sqrt{n} \times 40 \sqrt[6]{n} \\ 120 \sqrt[3]{ {n}^{a} } = 120 (\sqrt{n} \times \sqrt[6]{n} ) \\ \sqrt[3]{ {n}^{a} } = (\sqrt{n} \times \sqrt[6]{n} ) \\ {n}^{ \frac{a}{3} } = {n}^{ \frac{1}{2} } \times {n}^{ \frac{1}{6} } \\ \frac{a}{3} = \frac{1}{2} + \frac{1}{6} \\ \frac{a}{3} = \frac{4}{6} \\ 6a = 12 \\ a = \frac{12}{6} = 2 \\ \underline{ \boxed{a = 2 \:}} \\ \\ \: \underline{ \boxed{solution \: to \: \boxed{b}}} \\ 27 \sqrt[4]{ {n}^{b} } = 3 \sqrt{n} (9 \sqrt[4]{n} ) \\ 27 \sqrt[4]{ {n}^{b} } = 3 \sqrt{n} \times 9 \sqrt[4]{n} \\ 27 \sqrt[4]{ {n}^{b} } = 27 (\sqrt{n} \times \sqrt[4]{n} ) \\ \sqrt[4]{ {n}^{b} } = (\sqrt{n} \times \sqrt[4]{n} ) \\ {n}^{ \frac{b}{4} } = {n}^{ \frac{1}{2} } \times {n}^{ \frac{1}{4} } \\ \frac{b}{4} = \frac{1}{2} + \frac{1}{4} \\ \frac{b}{4} = \frac{3}{4} \\ 4b = 12 \\ b = \frac{12}{4} = 3 \\ \underline{ \boxed{b = 3 \:}} \\ hence \to : \\ \underline{ \underline{\boxed{a = 2 \:}} \underline{ \boxed{b = 3 \:}}}$