how do i calculate the power x


by chemart
Tags: power
chemart
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#1
Mar15-08, 03:39 PM
P: 6
[tex]2^x=128[/tex]
[tex]x=?[/tex]

how do i calculate power x from 128 and 2?
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John Creighto
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#2
Mar15-08, 04:18 PM
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Quote Quote by chemart View Post
[tex]2^x=128[/tex]
[tex]x=?[/tex]

how do i calculate power x from 128 and 2?
You take the log of both sides:

log(2^X)=log(128)
Then you use the property
that log(a^b)=b log(a)

The remaining steps should be apparent.
mgb_phys
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#3
Mar15-08, 04:20 PM
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Or quicker, just calculate 2*2*2.... until you get 128!

ice109
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#4
Mar15-08, 06:11 PM
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how do i calculate the power x


Quote Quote by mgb_phys View Post
Or quicker, just calculate 2*2*2.... until you get 128!
that's not quicker...
nicksauce
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#5
Mar15-08, 06:42 PM
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Quote Quote by ice109 View Post
that's not quicker...
Depends how far away your calculator is.
kurt.physics
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#6
Mar15-08, 07:07 PM
P: 100
Quote Quote by chemart View Post
[tex]2^x=128[/tex]
[tex]x=?[/tex]

how do i calculate power x from 128 and 2?
2x = 128

the first thing you have to do is get both of the bases the same so in 2x, x is the exponent (or power if you like) and 2 is the base. So what you have to do is get 128 with a base of 2. So,

128 = 2 * 2 * 2 * 2 * 2 * 2 * 2

128 = 27

There for,

2x = 27

So,

x = 7

This way may seem long, but take the example of 22(x+1) = 1024

1024 = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2
= 210


22(x+1) = 210

2(x+1) = 10

2x + 2 = 10

2x = 8

x = 4
sutupidmath
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#7
Mar15-08, 07:30 PM
P: 1,635
there are also other cases where you have to take logs of both parts, like it initially was suggested. Say you have:

[tex]2^{x+1}=35[/tex] you defenitely cannot express 35 as a power of 2, so in these cases you need to take the log of both parts, and choose a base that will be easier to work with.
[tex]log_a 2^{x+1}=log35=>(x+1)log_a 2=log_a35=>x+1=\frac{log_a35}{log_a2}[/tex]
mgb_phys
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#8
Mar15-08, 07:57 PM
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Quote Quote by ice109 View Post
that's not quicker...
Yes it is -
1, you can do it in your head in the time it takes to find the log button on the calculator

or if you happen to be an android
2, multiplying by 2 is a single shift instruction in a register which you can do in a single clock cycle, it would take at least 7 clock cycles to load the instruction into the FPU and get the answer back - especially if you have a long pipeline.
symbolipoint
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#9
Mar15-08, 10:06 PM
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Too Simple! find the prime factorization of 128. Now, what is the exponent?

128 = 2 * 64 = 2 * 2*2*2 * 2*2*2

Count the factors of 2 in the factorization of 128.
Gib Z
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#10
Mar17-08, 06:52 AM
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mgb_phys way is definitely quicker. Note when we take logs of both sides, all we achieve is [itex]x= log_2 (128)[/itex], and to actually evaluate that we must work out [itex]2^7 = 128[/itex] anyway.


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