cube root of 7 multiplied by the square root of 7 over sixth root of 7 to the power of 5
Solve.. Please help!!!

Question

cube root of 7 multiplied by the square root of 7 over sixth root of 7 to the power of 5
Solve.. Please help!!!

anonymous · Answer

$$^3\sqrt7\sqrt7/^6\sqrt7^5$$

hartnn · Answer

lets first use this : 
$\huge \sqrt[n]x = x^{1/n}$

so what about 
$\large \sqrt[3]7 = ... ?$

anonymous · Answer

would it be $$7^1/3$$

anonymous · Answer

1/3 is supposed to be in exponent form

hartnn · Answer

yes, i know :) 

then you can also answer me 
$\sqrt[6]7^5  = ... \ \sqrt 7 = ...$

anonymous · Answer

$$^6\sqrt7^5 = 7^5/6$$ and 
$$\sqrt7$$ stays the same right?
I am sorry, but I am not really good at this....

hartnn · Answer

you're good! 
both of your answers are correct :)
and for 
$\sqrt 7$
it becomes 
$\Large 7^{1/2}$

anonymous · Answer

oh ya....

hartnn · Answer

now moving to next rule : 
$\huge x^m x^n = x^{m+n}$

so
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