x√x
= √x√x√x
= √(x)^3
= x^(3/2)
Remember the laws of exponents (indices = plural of index)
When multiplying powers of the same base (here, the base is "x"), you add the exponents.
x^3 * x^2 = xxx * xx = xxxxx = x^5
x^3 * x^2 = x^(3+2) = x^5
When dividing, subtract:
x^7 / x^3 = xxxxxxx / xxx = xxxx = x^4
x^7 / x^3 = x^(7-3) = x^4
This is how you get to x^0 = 1
1 = something over itself, for example
1 = x^3 / x^3 = x^(3-3) = x^0
Fractions = negative exponents
1 / x^2 = x^0 / x^2 = x^(0-2) = x^(-2)
x by itself is the same as x^1 (we simply don't bother writing the 1)
-----
Roots?
The "square root" of a number is a value which, when multiplied by itself (when "squared") gives you back the original number.
The square root of 49 is 7, because 7 * 7 = 7^2 = 49
What is the square root of "x"?
What value, when it is squared, will give us "x^1"
Using exponents, we can write
x^? * x^? = x^1
x^? being the "square root" of x
Using the law of exponents, we have
x^(?+?) = x^1
which requires
? + ? = 1
which forces us to use ? = 1/2
---
Putting it all together:
x√x = x * √x = x^1 * x^(1/2) = x^(1 + 1/2) = x^(3/2)
Cube root of x = x^(1/3)
Seventh root of x = x^(1/7)
and so on.
The square of the cube root = x^(1/3) * x^(1/3) = x^(2/3)
The multiplicative inverse (fraction) of the square of the cube...
1/[x^(1/3) * x^(1/3)] = 1/[x^(2/3)] = x^(-2/3)
The fun is endless.
√x =x^(1/2)
x √x = x x^(1/2) = x^1 x^(1/2) = x^(1+1/2) = x^(3/2)
Solution --
Law of indices is :-
x^m X x^n = x^(m + n)
x^1 X x^(1/2) = x^(3/2)
Recall that: x^a * x^b = x^(a+b)
x√x = x^1 * x^(1/2) = x^(1+1/2) = x^(3/2)
x^1 * x^(1/2)= x^(1 + 1/2)= x^(3/2)
because it is 1+1/2 or (2+1)/2= 3/2, so xrt x= x*3/2, voila
x is x^1
Sqrt of x is x^(1/2)
Add exponents when terms are multiplied.
1+(1/2)=3/2
Yes
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Answers & Comments
x√x
= √x√x√x
= √(x)^3
= x^(3/2)
Remember the laws of exponents (indices = plural of index)
When multiplying powers of the same base (here, the base is "x"), you add the exponents.
x^3 * x^2 = xxx * xx = xxxxx = x^5
x^3 * x^2 = x^(3+2) = x^5
When dividing, subtract:
x^7 / x^3 = xxxxxxx / xxx = xxxx = x^4
x^7 / x^3 = x^(7-3) = x^4
This is how you get to x^0 = 1
1 = something over itself, for example
1 = x^3 / x^3 = x^(3-3) = x^0
Fractions = negative exponents
1 / x^2 = x^0 / x^2 = x^(0-2) = x^(-2)
x by itself is the same as x^1 (we simply don't bother writing the 1)
-----
Roots?
The "square root" of a number is a value which, when multiplied by itself (when "squared") gives you back the original number.
The square root of 49 is 7, because 7 * 7 = 7^2 = 49
What is the square root of "x"?
What value, when it is squared, will give us "x^1"
Using exponents, we can write
x^? * x^? = x^1
x^? being the "square root" of x
Using the law of exponents, we have
x^(?+?) = x^1
which requires
? + ? = 1
which forces us to use ? = 1/2
---
Putting it all together:
x√x = x * √x = x^1 * x^(1/2) = x^(1 + 1/2) = x^(3/2)
---
Cube root of x = x^(1/3)
Seventh root of x = x^(1/7)
and so on.
The square of the cube root = x^(1/3) * x^(1/3) = x^(2/3)
The multiplicative inverse (fraction) of the square of the cube...
1/[x^(1/3) * x^(1/3)] = 1/[x^(2/3)] = x^(-2/3)
The fun is endless.
√x =x^(1/2)
x √x = x x^(1/2) = x^1 x^(1/2) = x^(1+1/2) = x^(3/2)
Solution --
Law of indices is :-
x^m X x^n = x^(m + n)
x^1 X x^(1/2) = x^(3/2)
Recall that: x^a * x^b = x^(a+b)
x√x = x^1 * x^(1/2) = x^(1+1/2) = x^(3/2)
x^1 * x^(1/2)= x^(1 + 1/2)= x^(3/2)
because it is 1+1/2 or (2+1)/2= 3/2, so xrt x= x*3/2, voila
x is x^1
Sqrt of x is x^(1/2)
Add exponents when terms are multiplied.
1+(1/2)=3/2
Yes