How do I solve 2^(x) < 4?

dexter

I have some inequality 2^x < 4, first I look at the region x greater than or equal to 0 which gives me x*ln2<ln4 => x<ln2/ln4. Then I consider x<0 which is where I get stuck. I thought of a few things, like if x is negative, then ln(2^x) will be a negative value overall, or maybe I can bring the x down the front and get -x*ln2<ln4 which is x>-ln2/ln4.

I know the final solution is x<ln2/ln4, so what am I missing? I'm looking for a way to handle this fully.

Irridan

I believe that you are overthinking it some. If you just look at the first equation which you say you formulated for the case when x is >0, x<ln(2)/ln(4).

That will hold when x is negative as well. ln(2)/ln(4) is not a very large number, but it is always larger than any negative number!

MrTLicious

Try plotting 2^x.

ThunderSaid

Take a look at this explanation:
purplemath.com/modules/solvelog2.htm
I think it's helpful.

Also, I think you might have a misunderstanding about what negative exponents mean.
2^(-x) = 1/(2^x)
So, for example, 2^(-2) = 1/4, not -4

Dunadan019

the correct answer is x<ln4/ln2

ln4/ln2 is 2

2^x<4

2^2=4

a5ehren

1) As an EE, seeing all you guys use ln with 2^x numbers is making my brain hurt

2) But yeah, considering we're dealing with 2^x here, -x is always just going to be some fraction that is always less than 4. Given the way exponents work with negative numbers, you don't have to treat it any different.

The solution here is to use log2 on both sides - then you get this:

log2(2^x) < log2(4)

x < 2

So:
log2(2^-5) < log2(4)

-5 < 2, which is true.

Using ln gets you to the same answer in a less efficient way:
ln(2^x) < ln(4)

x*ln(2) < ln(4)

x < ln(4)/ln(2)

x < 2

The negative numbers work because -1 * ln(2) == ln(2^-1) == ln(0.5) == -0.693...

Heffling

dexter wrote: »

I have some inequality 2^x < 4, first I look at the region x greater than or equal to 0 which gives me x*ln2<ln4 => x<ln2/ln4. Then I consider x<0 which is where I get stuck. I thought of a few things, like if x is negative, then ln(2^x) will be a negative value overall, or maybe I can bring the x down the front and get -x*ln2<ln4 which is x>-ln2/ln4.

I know the final solution is x<ln2/ln4, so what am I missing? I'm looking for a way to handle this fully.

2^x < 4

Take the natural log of both sides:

ln(2^x) < ln(4)

Factor x out:

x * ln(2) < ln(4)

Divide both sides by ln (2):

x < (ln(4)) / (ln(2))

Convert 4 to 2^2:

x < (ln(2^2)) / (ln(2))

Factor 2 out:

x < 2 * ln(2) / ln(2)

Replace ln(2)/ln(2) with 1:

x < 2 * 1

Simplify:

x < 2


Notes:
ln(2^x) = x * ln(2)
ln(2^2) = 2 * ln(2)
Any number raised to a negative value is the same as one over that number to a positive value. So 2^-2 is the same as 1/(2^2). Either case equals 1/4.

Somehow you swapped your 2 and your 4 when dividing by ln(2) in your example.

dexter

Thanks guys, I got it I was over thinking the idea that if I had ln2^(-x) and x>0, then the entire expression would be negative.

Thanks again.

Rend

Yeah the natural log in this case is hilariously complex. You can simplify it immensely by just using log2.

This is actually a super important point: If you understand the things you are working with (what is log? what is log2? what is logx? what is the natural log?) at a deeper-than-surface level, you can start manipulating your normal process flow to benefit the problem at hand.

log2 is more appropriate than ln because log2 simplifies the equation, where ln adds additional steps.