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You need to remember the relation that exists between a rational exponent and a radical, such that:
x^(m/n) = root(n)(x^m)
Hence,the radical symbol can be converted into a rational exponent, whose denominator represents the order of radical.
You may start with the following exponent operation, such that:
x^(m/n) = (x^(1/n))^m
You also can replace y for x^(1/n) such that:
y = x^(1/n)
Raising both sides to nth power, you can get eliminate the rational exponent, such that:
`y^n = (x^(1/n))^n => x = y^n`
Hence, evaluating the nth root of x yields x^(1/n).
You may consider the following example, such that:
`root(3)(16) = 16^(1/3)`
You may replace `4^2` or `2^4` for `16` , such that:
`root(3)(16) = root(3)(4^2) = 16^(1/3) =(4^2)^(1/3) = 4^(2/3)`
`root(3)(16) = root(3)(2^4) = (2^4)^(1/3) = 2^(4/3)`
You may notice that the order of radical (3) becomes the denominator of rational exponent and the numerator is the power(exponent) of the number under the radical.
Hence, you may convert the radical into a rational exponent, such that:`root(3)(16) = 4^(2/3) =2^(4/3)`
Hence, if you need to perform the conversion between radical in rational exponent or conversely, you need to use the formula `x^(m/n) = root(n)(x^m).`
You have to think about what an esponent means:
If we say: `a^n=b` it means:
`axxa xxa xxaxxaxxa.....xxa` (n times) `=b`
Now often te problem is to find, given a, the number b so that:
`a^n=b` (exponential equation)
On the other side we can have the inverse problem, that means:
given `b` , what is the a so that: `a^n=b ?``` (Root problem)
I. e if we have a square area `b` if we wanna come to know the side ha ve to know what is the number `a` so that `a^2=b`
Or if we have cube volume `b` ,and wanna to come know the side of the cube, it means wonder: what is the number `a` so that `a^3=b` ?
To solve root problem `a^n=b` that is to find the symbolic number `root(n)(b)`
You know , from exponential rules that product of power of the same base is the sum of the power of base itself:
`a^n xx a^m= a^(n+m)`
Conseguently power of a power fo a bse si the product of power:
Concering on the equation `a^n=b` we can wonder:
"can we name `a` as `b^(1/n)` ?"
Isn't only a useless different expression by `root(n)(b)` , but as ratio, allows us to operate on exponents as product of ratio, making more easy to find a result.
Indeed we can easiest find a "name" at expression: `a^(m/n)`
`a^(m/n)= (a^(1/n))^m` than `(root(n)(a))^m`
Now we know, that if `x=y` , of course `x^n=y^n`
So if we put: `` `a=b^(1/n)` we have too: `a^n=(b^(1/n))^n=`
`=b^(1/n xx n)=b^1=b`
That proves `b^(1/n)=a`
``What does is it usefull?
We have rigth seen that `(root(n)(a))^m=` `a^(m/n)` ``
On the other side `a^(m/n)=a^(m xx 1/n)=(a^m)^(1/n)`
TO BE CONTIUNION......
So, `(root(n)(a))^m= a^(m/n)=(a^m)^(1/n)=root(n)(a^m) `
From exponential rules we now also : `x^n xx y^n= (xy)^n` . (1)
Suppose to find: `root (3)(9 xx 96 xx 98 xx 7)`
Using merely roots system of calculation we have to find the root:
Instead using ratio meaning:
`root(3)(9 xx 96 xx 98 xx 7)= root(3)(3^2 xx 3 xx 2^5 xx 2 xx 7^2 xx7)=`
Setting all bases power togheter from (1):
`=root(3)(3^3 xx 2^6 xx7^3)=` `(3^3 xx 2^6 xx 7^3)^(1/3)=`
Using (1) again:
`=(3^3)^(1/3) xx (2^6)^(1/3) xx (7^3)^(1/3)=`
Using again ratio meaning:
`=3^(3xx 1/3) xx 2^( 6 xx 1/3) xx 7^(3xx 1/3)=` `3^1 xx 2^2 xx 7^1= 3xx2^2 xx 7=84`
Let you see that roots operation didn't compel us to make moltiplication and afterword find a root with hard calcutating, instead, allows us to divide operation in easiest steps in order to find the same result, (indeed `84^3= 592,704`
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