# Difficult Power Rule task (without calculator)

• Jul 20th 2010, 08:57 AM
kevin3000
Difficult Power Rule task (without calculator)
I really need help with this one.

• Jul 20th 2010, 10:18 AM
undefined
2^7 = 128. So...
• Jul 20th 2010, 12:26 PM
1005
Quote:

Originally Posted by undefined
2^7 = 128. So...

How does that help him find an estimation for 128^48 without using a computer?
$\displaystyle 128^{48}=(2^{14})^{24}=16384^{24}$

Or am I missing something? All I know is I wouldn't want to do 16384^14 by hand.
• Jul 20th 2010, 12:29 PM
undefined
Quote:

Originally Posted by 1005
How does that help him find an estimation for 128^48 without using a computer?
$\displaystyle 128^{48}=(2^{14})^{24}=16384^{24}$

Or am I missing something? All I know is I wouldn't want to do 16384^14 by hand.

All we need is an estimate. Why not take

$\displaystyle 128^{48}=16384^{24} \approx (10^4)^{24} = 10^{96}$

?
• Jul 20th 2010, 02:20 PM
Soroban
Hello, kevin3000!

This is a strange problem . . .

Quote:

Calculate an approximate value to $\displaystyle 128^{48}$
if you know that: $\displaystyle 2^{14} = 16,\!384$ . . I don't see how this is useful.

I would do it like this . . .

We have: .$\displaystyle N \;=\;128^{48} \:=\:\left(2^7\right)^{48} \:=\:2^{336}$

Since $\displaystyle 2^{10} \:=\:1024 \:\approx\:1000 \:=\:10^3$

we have: .$\displaystyle N \;=\;2^6\cdot2^{330} \;=\;2^6\cdot (2^{10})^{33}$

. . . . . . . .$\displaystyle N\;\approx\;2^6\cdot(10^3)^{33} \;=\;64\cdot10^{99}$

Therefore: .$\displaystyle N \;\approx\;6.4 \times 10^{100}$

• Jul 20th 2010, 02:45 PM
1005
undefined's approximation(10^96) is 5 order of magnitude off with 99.99% error (lol)

and Soroban's approximation is 1 order of magnitude off with 54.28% error.
• Jul 20th 2010, 03:11 PM
undefined
Quote:

Originally Posted by 1005
undefined's approximation(10^96) is 5 order of magnitude off with 99.99% error (lol)

and Soroban's approximation is 1 order of magnitude off with 54.28% error.

It's funny yes but I was merely following the instructions of the problem (or trying to do so, since the problem doesn't make much sense); given freedom, I would also use the approximation 2^10 $\displaystyle \approx$ 10^3.

Anyway if you go high enough, 2^10 $\displaystyle \approx$ 10^3 will give you 99.99% error as well.
• Jul 20th 2010, 05:37 PM
bjhopper
Estimates
[QUOTE=1005;539205]undefined's approximation(10^96) is 5 order of magnitude off with 99.99% error (lol)

and Soroban's approximation is 1 order of magnitude off with 54.28% error.[/QUOTE

Soroban's approx is not one magnitude off. It can be expressed as 0.64E101. Calculator answer is 1.4xE 101

bjh