Factorizing an 'i' in Exponent: Justified?

[Euler2718]

In the attached picture, how is it justified to factor out an 'i' in the exponent of the 2 and the 5?

 

[fresh_42]

Do you know what $2^i$ means?

 

[Euler2718]

Do you know what $2^i$ means?

Apparently I do not. To clarify, I'm not working with imaginary, "i" is just the variable they choose.

 

[fresh_42]

Apparently I do not. To clarify, I'm not working with imaginary, "i" is just the variable they choose.

I know. And with another letter: $2^n$ is simply a short form of multiplying $2$ with itself $n$ times.
$2^0 = 1$ (convention), $2^1 = 2, 2^2 = 2 \cdot 2 = 4, 2^3= 2 \cdot 2 \cdot 2 =8$ and so on.

 

[Euler2718]

I know. And with another letter: $2^n$ is simply a short form of multiplying $2$ with itself $n$ times.
$2^0 = 1$ (convention), $2^1 = 2, 2^2 = 2 \cdot 2 = 4, 2^3= 2 \cdot 2 \cdot 2 =8$ and so on.

Okay. But I'm still not sure on why they can factor out $\frac{2^{i}}{5^{i}}$ algebraically. I get that you can take the 2 out from the top and the 5 from the bottom but how do they get an exponent too?

 

[fresh_42]

Okay. But I'm still not sure on why they can factor out $\frac{2^{i}}{5^{i}}$ algebraically. I get that you can take the 2 out from the top and the 5 from the bottom but how do they get an exponent too?

That's why $(\frac{a}{b})^n = \frac{a}{b} \cdot ... \cdot \frac{a}{b} (n$ times $) = \frac{a \cdot ... \cdot a }{b \cdot ... \cdot b}$ each $n$ times.
$(\frac{2}{5})^3 = \frac{2}{5} \cdot \frac{2}{5} \cdot \frac{2}{5} = \frac{2 \cdot 2 \cdot 2}{5 \cdot 5 \cdot 5} = \frac{2^3}{5^3}$
And $2$ is contained in every single factor of the nominator and $5$ in every single factor of the denominator.

 

[Euler2718]

That's why $(\frac{a}{b})^n = \frac{a}{b} \cdot ... \cdot \frac{a}{b} (n$ times $) = \frac{a \cdot ... \cdot a }{b \cdot ... \cdot b}$ each $n$ times.
$(\frac{2}{5})^3 = \frac{2}{5} \cdot \frac{2}{5} \cdot \frac{2}{5} = \frac{2 \cdot 2 \cdot 2}{5 \cdot 5 \cdot 5} = \frac{2^3}{5^3}$
And $2$ is contained in every single factor of the nominator and $5$ in every single factor of the denominator.

Sorry, I still don't follow.

 

[fresh_42]

Sorry, I still don't follow.

$2 \cdot 4 \cdot 6 \cdot \cdot \cdot 2i = (2 \cdot 1) \cdot (2 \cdot 2) \cdot (2 \cdot 3) \cdot \cdot \cdot (2 \cdot i) = [ 2 \cdot 2 \cdot 2 \cdot \cdot \cdot (i$ times $) \cdot \cdot \cdot 2] \cdot [1 \cdot 2 \cdot 3 \cdot \cdot \cdot i] = 2^i \cdot i!$
and the same with $5$ in the denominator. Then $i!$ cancels out and $\frac{2^i}{5^i} = (\frac{2}{5})^i$ is left.

 

[Euler2718]

$2 \cdot 4 \cdot 6 \cdot \cdot \cdot 2i = (2 \cdot 1) \cdot (2 \cdot 2) \cdot (2 \cdot 3) \cdot \cdot \cdot (2 \cdot i) = [ 2 \cdot 2 \cdot 2 \cdot \cdot \cdot (i$ times $) \cdot \cdot \cdot 2] \cdot [1 \cdot 2 \cdot 3 \cdot \cdot \cdot i] = 2^i \cdot i!$
and the same with $5$ in the denominator. Then $i!$ cancels out and $\frac{2^i}{5^i} = (\frac{2}{5})^i$ is left.

I think I understand now! I feel pretty stupid lol. Much appreciated.