Question

can someone help me explain this (attachment)

Answer 1

sorry i don't know but try and look it up on WolframAlpha.com
good luck :]]]]

Answer 2

what grade are you in btw?

Answer 3

\[x^{p/q} = \sqrt[q]{x^p}\]

Answer 4

The definition of exponentiation begins with:
x^n=x.x.x....x (n times) where n is naturally a positive number.
This definition leads to (for n, m being two positive integers)
{A} (x^n).(x^m)=x^(n+m) since:
[x^n=x.x.x....x (n times)].[x^n=x.x.x....x (m times)]=x^n=x.x.x....x (n+m times)=x^)n+m)
And of course similarly: (x^n)/(x^m)=x^(n-m) PROVIDED n>m
Next we can see that: (x^n)^m=[x^n=x.x.x....x (n times)].[x^n=x.x.x....x (n times)]....[x^n=x.x.x....x (n times)] where the terms within the square bracket is repeated m times giving us [x^n=x.x.x....x (mn times)] and thus:
{B} (x^n)^m=x^(mn)


From here on, the idea is to EXTEND the definition to cover:
1. Negative values of n in x^n
2. Rational or Fractional values of n in x^n
3. Real (irrational + rational) values of n in x^n

While doing so we MUST preserve the properties {A} and {B} above.

So if we with to extend the definition of exponentiation to Rational number, we start with numbers of the form 1/p where p is a positive integer and apply Property B as follows:

Let x^(1/p)=y.
Now then, using Property B it should be that: y^p=(x^(1/p))^p=x^((1/p).p)=x^1=x
So we get x=y^p or in other words, y is the pth root of x. But then y was defined as x^(1/p). Hence:

x^(1/p)=pth root of x.

And so: x^(1/5)= 5th root of x.

I guess you can perhaps work ahead along this line to solve the other problems.

Good luck!

Answer 5

thanks! could you also help me with #52 and 53