Comparing 2^100 + 3^100 and 4^100

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[@GourangaPL]

why not just divide everything by 4^100?

[@jalma9643]

Interesting enough

[@ericlipps9459]

That's right. If you do that, you get (2/4)^100+(3/4)^100 on the left side and 1 on the right, and since 2/4 and 3/4 are less than 1, raising them to any positive nonzero power will always produce a sum less than 1. Therefore, the sum on the left side must be greater than the number on the right side.

[@goldfing5898]

@@ericlipps9459 Exactly, except that (1/2)^n + (3/4)^n < 1 only for n >= 2, not for any natural number.
For n = 1, you get 1/2 + 3/4 = 1 1/4 > 1.
But already for n = 2, you get 1/4 + 9/16 = 4/16 + 9/16 = 13/16 < 1.

[@maxhagenauer24]

​@ericlipps9459 Yes, you know each of those 2 fractions raised to a positive non zero power makes them smaller but how do you know their sum afterwards is still less than 1?

[@GourangaPL]

@@maxhagenauer24 i told before, (3/4)^100 < 3/4 and (1/2(^100 < 1/4 so sum of two things smaller than 3/4 and 1/4 must be smaller than 1

[@ManjulaMathew-wb3zn]

Here is a super easy method.
2^100 + 3^100 vs 4^100
4^50 + 3*3^99 vs 4*4^99
4^50 + 3*3^99 vs 4^99 + 3* 4^99
Each of the term on RHS is greater than corresponding term on LHS so LHS

[@ManjulaMathew-wb3zn]

In fact I managed to prove the following inequality too.
16^25 + 3*27^33

[@ProficiencyMusic]

It's easier to divide all sides by 2¹⁰⁰, you get 1+(3/2)¹⁰⁰ vs 2¹⁰⁰.

[@mystychief]

4^100=2^200=2*2^199=2^199+2^199 compare with 2^100+3^100. 2^199>2^100, so we now only have to prove that 2^199>3^100 (which is obviously the case). You can do that via the log (because log(x) is increasing you can do that). So compare 199log(2) with 100log(3). The first is definitively bigger.

[@yoav613]

Yaaay nice problem that i can solve in my head😊😊

[@iabervon]

I divided by 3^100, which gives 1+(2/3)^100 versus (4/3)^100. 2/3 is already less than 1, so the LHS is less than 2. 4/3 is greater than 1, so it goes up reasonably quickly with higher powers. 64/27 is enough to be more than 2, or 256/81 works if you've just memorized squares and powers of 2. Alternatively, 1+(2/3)^n is headed for less than 4/3, with 1+8/27 already less.
Now that I think of it, 2²+3²

[@SyberMath]

you can also compare 2 and (4/3)^100 which can be done with the Binomial Theorem after writing (4/3)^100 as (1+1/3)^100.

[@iabervon]

@@SyberMath That's really nice, since the second term (100/3) is clearly bigger than 2 but small enough to be easy to think about.

[@user-cd9dd1mx4n]

Note that both 2/4=1/2 and 3/4 are positive and strictly less than 1, raising both of them to a (high) power (like 100), will result values that are (very close) to zero. Adding them will still be (close to) zero. On the other hand, 4/4=1, raising it to 100, will still be 1. Hence 4^100 is the greater one.

[@hel13ay]

Divede by 4^100 and get (1/2)^k+(3/4)^k and 1 respectively. For k bigger than 2 the left side is less than 1. As exponent is monotonical this hold for every k bigger than 2. So, 4^100 is bigger.

[@GlorifiedTruth]

I changed 2^(100) to 4^50. Then I figured 3 followed by 100 zeroes plus 4 followed by 50 zeroes is clearly less than 4 followed by 100 zeroes.

[@actions-speak]

You can also consider the binomial expansion of (3+1)^100.

[@leif1075]

Or 2 plus 1^100 as I did right?

[@rakenzarnsworld2]

2^200

[@leif1075]

I dint see why anyone would breaknthe 3 into 3 7 though. Isnt that out of nowhere? Why not 3 to 98 or 96. It's totally arbitrary right .

[@SyberMath]

If you do it any other way, it won't work. 3 and 97 are two good numbers because 2*3^3 is 54, it's less than and pretty close to 4^3 which is multiplied by 4^97 to form 4^100. As a result that allows you to compare 2*3^100 and 4^100. I think you can also compare 2 and (4/3)^100 which can be done with the Binomial Theorem after writing (4/3)^100 as (1+1/3)^100.

[@devondevon4366]

4^100 is larger

[@user-cd9dd1mx4n]

I think
2²⁰⁰ + 3³⁰⁰ VS 4⁴⁰⁰ is a nice one to try. Hope you can so a video about it. I can also share my solution.

[@SyberMath]

Can you share your solution? Thanks