I need help

I need help with understanding the explanation behind a problem here

$p=\frac { { 4 }^{ 3^{ 2 } } }{ { 2 }^{ 3^{ 4 } } }$

Which of the following is correct about the above number P?

Here's the explanation Brilliant gave me.

$P\quad =\quad \frac { { 4 }^{ 9 } }{ { 2 }^{ 81 } } \quad =\quad \frac { { 2 }^{ 18 } }{ { 2 }^{ 81 } } \quad =\quad { 2 }^{ -63 }\quad =\quad \frac { 1 }{ 8 } ({ 2 }^{ 10 })^{ -6 }\quad \approx \quad \frac { 1 }{ 8 } (10^{ 3 })^{ -6 }\quad =\quad \quad \frac { { 10 }^{ -18 } }{ 8 } .\quad$

The answer: $P\quad <\quad { 10 }^{ -10 }\quad$

I particularly want help with how you would go about getting the 1/8* (2^10)^-6 in the simplified expression. All I really need is some clarification in the explanation. Thanks ahead of time.

#Algebra

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Comments

$2^{-63} = (2^{-3})(2^{10})(2^{-6})$ since we add exponents when multiplying with like bases.

$2^{-3} = \frac{1}{2^3} = \frac{1}{8}$

Hope that helps.

David Stiff - 9 months, 4 weeks ago

That brings up another question I have, How did you get 2^-3 * 2^10 * 2^-6? I'm lost on how you got that.

Brian Harahus - 9 months, 4 weeks ago

Oops. Sorry, I think I made a small mistake. I should have said $2^{-63} = (2^{-3})(2^{-60})$ (since we can add the exponents), and then we can further expand $2^{-60}$ to get $(2^{-3})(2^{10})^{-6}$ . Does that make sense?

David Stiff - 9 months, 3 weeks ago

@David Stiff – Yes, thank you, that was a huge help. I was getting kind of confused there for a minute

Brian Harahus - 9 months, 3 weeks ago

@Brian Harahus – Sorry about that. :) Glad it helped.

David Stiff - 9 months, 3 weeks ago

Because $2^{10}=1024\approx10^3\implies(2^{10})^{-6}\approx(10^3)^{-6}$

A Former Brilliant Member - 9 months, 3 weeks ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$