Further roots

Further roots to explain `x^(a/b)`

Always split the power into a root and a power.

`x^(a/b)=x^((a\ \m\u\l\t\i\p\l\i\e\d\ \by\ \1/b))=(x^a)^(1/b)=rootb(x^a)` 

Or

`x^(a/b)=x^((1/b\ \m\u\l\t\i\p\l\i\e\d\ \by\ \a))=(x^(1/b))^a=(rootbx)^a`

Just remember:

Example 1

What is `10^(2/10)` ?

`10^(2/10) =(\root10\10)^2` and reads as

Work out what we multiply by itself `10` times to get `10`, then multiply the answer by itself `2` times. 

In this example

The number we multiply by itself `10` times to get `10=1.259` (to find this see logarithms)

Then multiply `1.259` by itself `2` times.

                     `1.259xx1.259=1.585`

So `10^(2/10)=1.585`

Example 2

Break any fraction up and use simple numbers  

`8^(2/3)=8^((2)times(1/3)`

This can be rewritten as

`8^((2)\times(1/3))=(8^2)^(1/3)=(64)^(1/3)=root3\64=4`

This is worded as multiply `8` by itself `2` times, then workout what we multiply by itself `3` times to get this answer. The answer is `4`.

 Or (working either way)

 `8^((1/3)times(2))=(8^(1/3))^2=(root3(8))^2=(2)^2=4`

This is worded as: work out what we multiply by itself `3` times to get `8`, then multiply the answer by itself `2` times.

So  `8^(2/3)=(root3\8)^2`

Example 3

Simplify `root4\(x^4)`

As follows:

`root4\(x^4)` is the same as `x^(4/4)`

Therefore `x^1`

Answer: The simplification of `root4\(x^4)=x`

Example 4

What is `27^(4/3)` ?

`27^(4/3)=27^(4times(1/3))=root3((27^4))=root3((531441))=81`

Or we could do

`27^(4/3)=27^((1/3)times4)=(root3\27)^4=(3)^4=81`

The answer is the same but in this case the second calculation was much easier.