Can someone clarify a problem for me? It's a function.

Question

angel_kitty12 · Answer

$$f(f(x)); f(x)=2x^\frac{ 1 }{ 5 }$$

angel_kitty12 · Answer

I want to make sure I'm doing this correctly.

zepdrix · Answer

You're trying to figure out f(f(x))?
Where is your work? What did you end up with? :)

angel_kitty12 · Answer

okay I got this:

f(2x^1/5)= 2(2x^1/5)^1/5
                
=2(2^1/5x^1/25)
Then I have to multiply it out.

angel_kitty12 · Answer

so i did it wrong at first but someone told me to multiply it like this 

2^1/1 * 2^1/5
2^1/1*5
2^5/5*2^1/5

to get 
2^6/5

angel_kitty12 · Answer

And my final answer was 2^6/5x^1/25 but i want to know why i have to multiply the 2s as i did

zepdrix · Answer

Ok looks great :)
Because you're applying the exponent addition rule,$$\large\rm x^a\cdot x^b=x^{a+b}$$So for our 2's,$$\large\rm 2\cdot 2^{1/5}=2^1\cdot 2^{1/5}=2^{1+\frac{1}{5}}$$And then you need a common denominator within your exponent in order to add those values together, ya?

angel_kitty12 · Answer

Question, why don't I multiply the 2's?

zepdrix · Answer

Hmm let's look at an example to see why.
$\large\rm 2\cdot 2^2$
This is a nice simple number to work out.
We know that 2^2 is 4,
so 2 times 4 is 8.

Ok let's try doing it using the other approach,
using our exponent addition rule,
$\large\rm 2^1\cdot 2^2=2^{1+2}=2^3=8$
So it gave us the correct result.

Notice that if we had multiplied the 2's,
$\large\rm 2^1\cdot2^2\ne 4^{1+2}$
it gives us a completely different answer, a wrong answer actually.

zepdrix · Answer

`Why don't I multiply the 2's?`
I guess if I can try to say it in a more clear way,
it's because of what exponent actually means.

You're just multiplying a bunch of 2's together.
It shouldn't turn into another number if you're putting a bunch more 2's into that group.

angel_kitty12 · Answer

Oh so we wouldn't multiply the bases if they're the same but say if it was 3^1 *2^1/2 you would multiply to get 6^3/2

angel_kitty12 · Answer

Oh, okay that makes sense.

zepdrix · Answer

If we had something like $\large\rm 3^1\cdot 2^{1/2}$
You wouldn't do anything with it.
You would leave it like that.

You can't apply your exponent addition rule because the bases are different.
And you can't directly apply multiplication,
because exponent comes before multiplication in our PEMDAS, ya? :)

angel_kitty12 · Answer

Oh alright, I understand. Thank you so much!

zepdrix · Answer

You COULD do something fancy like this though.
You could rewrite 3 as the square root of 9,

$\large\rm 3\cdot 2^{1/2}=9^{1/2}\cdot 2^{1/2}$
Which could be written as
$\large\rm (9\cdot2)^{1/2}=18^{1/2}$
But that's a little more fancy I guess :))