How do i calculate the power x

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Discussion Overview

The discussion revolves around calculating the exponent \( x \) in the equation \( 2^x = 128 \). Participants explore various methods for solving this problem, including logarithmic approaches and direct multiplication, while also considering alternative cases where logarithms are necessary.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants suggest taking the logarithm of both sides to solve for \( x \), using the property \( \log(a^b) = b \log(a) \).
  • Others propose a more straightforward method of multiplying \( 2 \) repeatedly until reaching \( 128 \), although this is contested as not being quicker.
  • A participant points out that in cases where the number cannot be expressed as a power of \( 2 \), such as \( 2^{x+1} = 35 \), logarithmic methods become necessary.
  • One participant argues that calculating the prime factorization of \( 128 \) is a valid approach, leading to counting the factors of \( 2 \).
  • Another participant notes that while logarithmic methods yield \( x = \log_2(128) \), evaluating this still requires recognizing that \( 2^7 = 128 \).

Areas of Agreement / Disagreement

Participants express differing opinions on the efficiency of various methods for calculating \( x \). There is no consensus on which method is superior, and multiple approaches are presented without resolution.

Contextual Notes

Some methods discussed depend on the ability to express numbers as powers of \( 2 \), while others require logarithmic calculations, indicating a limitation based on the specific values involved.

chemart
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2^x=128
x=?

how do i calculate power x from 128 and 2?
 
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chemart said:
2^x=128
x=?

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

that's not quicker...
 
ice109 said:
that's not quicker...

Depends how far away your calculator is.
 
there are also other cases where you have to take logs of both parts, like it initially was suggested. Say you have:

2^{x+1}=35 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.
log_a 2^{x+1}=log35=>(x+1)log_a 2=log_a35=>x+1=\frac{log_a35}{log_a2}
 
ice109 said:
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.
 
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.
 
mgb_phys way is definitely quicker. Note when we take logs of both sides, all we achieve is x= log_2 (128), and to actually evaluate that we must work out 2^7 = 128 anyway.
 

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