# How do i calculate the power x

by chemart
Tags: power
 P: 6 $$2^x=128$$ $$x=?$$ how do i calculate power x from 128 and 2?
P: 813
 Quote by chemart $$2^x=128$$ $$x=?$$ 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.
 Sci Advisor HW Helper P: 8,954 Or quicker, just calculate 2*2*2.... until you get 128!
P: 1,705
How do i calculate the power x

 Quote by mgb_phys Or quicker, just calculate 2*2*2.... until you get 128!
that's not quicker...
HW Helper
P: 1,275
 Quote by ice109 that's not quicker...
Depends how far away your calculator is.
P: 100
 Quote by chemart $$2^x=128$$ $$x=?$$ 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
 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: $$2^{x+1}=35$$ 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. $$log_a 2^{x+1}=log35=>(x+1)log_a 2=log_a35=>x+1=\frac{log_a35}{log_a2}$$
 HW Helper P: 3,352 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.