Comparing 2^sqrt(7) and 3^{13/8}

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[mohamedhamouch]

Hi, if you use the calculator to compare 2^21 and 3^13 (at 5:20), then you can use it from the beginning to compare the whole expressions !

[megalomorph]

The solution would be a lot more motivated if the first step was to take both sides to the 8th power, as then the right hand side is an integer. This gives the problem of approximating 8sqrt(7), which squares to be 448, which is just above 21^2. Thus, the left hand side is between 2^21 and 2^22 (closer to the former), and you are comparing those to 3^13.

This approach doesn’t require you to know the answer in advance, and doesn’t require any unmotivated approximations. It is the same calculations, just presented in a less ad hoc manner.

[pwmiles]

Or use 3^5=243<256=2^8, 3^(5/8)<2. Also (13/5)^2=169/25<7, 13/5<sqrt(7). So 3^(13/8)=(3^(5/8))^(13/5)<2^(13/5)<2^sqrt(7), done. (I found this by beginning with close powers of 2 and 3)

[ジョン永遠]

Simply we use common logarithm. Let A=2^√7 and B=3^{13/8}.
We want to show a=log A>b=log B to have A>B. a=√7log2, b=(13/8)log3.
First √7>2.6 and log2>0.3 (∵ 2²1.3²=4x1.69<4x1.7=6.8<7and 2¹⁰>10³.)
∴ a>2.6x0.3=0.78.
Next log3<0.48 (∵ 3⁵=243<250=10³/2², so 5log3<3-2log2<3-2x0.3=2.4
⇒ 10log3<4.8.) ∴ b<(13/8)x0.48=13x0.06=0.78. Now we get a>0.78>b.

[strikerstone]

2^√7 3^(13/8)
2^2.6 remember 2.6>√7
2^(13/5)
2^(1/5) 3^(1/8)
raising to ^40 we get
2^8 3^5
256 243

if 2.6 >√7 and raised to 2.6 is more than that of other term so ofc 2^√7 is greater

[MichaelRothwell]

I followed essentially the same approach as SyberMath, but only after finding which looked bigger by estimation. By coincidence I estimated √7 as 21/8, know that 2^21≈2M, 3^7=2187, 3^14≈4.8M, 3^13≈1.6M. By replacing the ≈ by appropriate inequalities we get a proof.

[robertveith]

Know/recall that sqrt(3) ~ 2.65, ln(2) ~ 0.69, and ln(3) ~ 1.1 for the following method. Have the
left-hand side raised to the sqrt(7) so that it becomes 2^7. Have the right-hand side raised to the
decimal approximation 2.65. That is, 3^(1.625*2.65) ~ 3^(4.31). Take natural logs of both sides:
7*ln(2) versus 4.31*ln(3) 7(0.69) versus 4.31(1.1) 4.83 > 4.741.
So, 2^sqrt(7) > 3^(13/8). *More in favor of this is that 0.69 < ln(2), combined with 1.1 > ln(3) and 2.65 > sqrt(7).*

[GerhartHansen]

you can also solve like this: 
26^2 = 676 < 700 => 26^2/10^2 < 700/10^2 => (13 / 5)^2 < 7 => 13 / 5 < √7 => 2^(13/5) < 2^√7
256 > 243 => 2^8 > 3^5 => 2^(1/5) > 3 ^ (1/8) => 2^(13/5) > 3 ^ (13/8)
combining both lines we get the result 2^√7 > 3 ^ (13/8)

[TedHopp]

You sort of give the answer away near the beginning when you focus on finding only a lower bound for ✓7. Without knowing the answer in advance, one would be interested in both an upper and lower bound.

[Drk]

Bravo! Me gustó el saludo final, reconociendo a tus seguidores hispanos :) A little sugestion: 2^21 AND 3^13 are numbers too big. It was More didactical this (specially if we cannot use calculators): (2^21)/(3^13) = [((2^11)^2)/2] / [((3^7)^2)/3]= (3/2)*(2048/2187)^2. Taking 2048/2187 close enough to 1, then (2^21)/(3^13) is aprox 3/2>1. Later, demostration would keep idem :) Ps: sorry for my grammar fails. Greetings from Argentina, nice video!

[ahmadmazbouh]

8:20 about this new problem, is "by previous knowledge" a good argument for solving maths? if so i'd solve it in √π seconds

[carloshuertas]

Another great explanation, SyberMath!

[aliscais]

I think I have a bit easier approach:
2.6^2 = 6.76 < 7
2.6 < sqrt(7) or 13/5 < sqrt(7)
so 2^(sqrt(7)) > 2^(13/5)
let's compare 2^(13/5) and 3^(13/8)
it will be the same as to compare 2^8 and 3^5
2^8 = 256, 3^5 = 243 => 2^8 > 3^5 => 2^(sqrt(7)) > 2^(13/5) > 3^(13/8)

[nirajkumarverma]

Yes, really enjoyed it as I learnt new thing.👍👍👍

[giuseppemalaguti]

2^rad7>2^2, 6=2^(13/5)>3^(13/8)....qiesto perché 2^(1/5)>3^(1/8), essendo, elevando alla 40, 2^8>3^5.…questo perché 256>243....quindi in sintesi 2^(rad7)>3^(13/8)

[usagi.daisuki.]

What if we use log to solve this question?


0.796446>0.7752875 ∴2^√7>3^(13/8)
So we can get the answer is that 2^√7 is greater.

[abubakkarjutt]

Nice explanation... Sir I have a question about right angle triangle we frequently use a principle that if two or three angles of a triangle are equal then their sides will also be equal or vice versa... But if we go through the derivation of trignometric ratios of common angles like 30⁰, 60⁰ we take the side opposite to 30 as 1 side opposite to 60 as √3 and side opposite to 90 as 2... Don't you think so that if 60 is double of 30 then this ratio must also exist in their sides but if the ratio exists then Pythagorean theorem is not satisfied

[sahilagarwal]

Simply log both sides amd put the values would be an easy approach

[darknight]

Whats the point of solving it that way if you already used a calculator several times?

[GourangaPL]

you can estimate that sqrt(7) is between 13/5 and 11/4 using the formula i found few years ago