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## Main Question or Discussion Point

[tex]2^x=128[/tex]

[tex]x=?[/tex]

how do i calculate power x from 128 and 2?

[tex]x=?[/tex]

how do i calculate power x from 128 and 2?

- Thread starter chemart
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[tex]2^x=128[/tex]

[tex]x=?[/tex]

how do i calculate power x from 128 and 2?

[tex]x=?[/tex]

how do i calculate power x from 128 and 2?

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You take the log of both sides:[tex]2^x=128[/tex]

[tex]x=?[/tex]

how do i calculate power x from 128 and 2?

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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Or quicker, just calculate 2*2*2.... until you get 128!

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that's not quicker...Or quicker, just calculate 2*2*2.... until you get 128!

nicksauce

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Depends how far away your calculator is.that's not quicker...

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2[tex]2^x=128[/tex]

[tex]x=?[/tex]

how do i calculate power x from 128 and 2?

the first thing you have to do is get both of the bases the same so in 2

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

128 = 2

There for,

2

So,

x = 7

This way may seem long, but take the example of 2

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

= 2

2

2(x+1) = 10

2x + 2 = 10

2x = 8

x = 4

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[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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Yes it is -that's not quicker...

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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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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