Indices and roots

[Algebraneophyte]

Hi Guys.



I got question 7 wrong as per picture, could you please show me how to solve this to the answer? Which is 9√X.







The numbers will be different on my next attempt, so I would like reverse engineer this one to understand it and score 100% soon.
(Question 8 i simply forgot to root the 25 in addition to the x^4.)



Thank you for taking the time

 

[lex]

[MATH]\frac{\sqrt[3]{729x^2}}{\sqrt[6]{x}}=\frac{{(729x^2)}^\frac{1}{3}}{x^\frac{1}{6}}[/MATH]
[MATH]=\frac{729^\frac{1}{3}x^\frac{2}{3} }{x^\frac{1}{6}}[/MATH]
[MATH]=9 x^{(\frac{2}{3}-\frac{1}{6})}[/MATH]
[MATH]=9x^\frac{1}{2}[/MATH]
[MATH]=9\sqrt{x}[/MATH]

 

[HallsofIvy]

9*9= 81 and 9*81= 729 so \(\displaystyle 9^3= 729\) and so \(\displaystyle 729^{1/3}= 9\).

\(\displaystyle \frac{\sqrt[3]{729x^2}}{\sqrt[6]{x}}= \frac{9 x^{2/3}}{x^{1/6}}= 9x^{2/3- 1/6}= 9x^{4/6- 1/6}= 9x^{3/6}= 9x^{1/2}= 9\sqrt{x}\).