Let's Compare 2⁸³ and 7³¹

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[@trumpetbob15]

I used a slightly different approach. 2^11 = 2048 < 7^4 = 2401. Raise both sides by power 8 so we get 2^88 < 7^32. This can also be written as 2^83*2^5 < 7^31 * 7. If we were to divide both sides by 7, we get one side of our final result of 7^31 but we are still multiplying 2^83 by 32/7, which is greater than 1 - meaning 2^83 is less than 7^31.

[@MichaelRothwell1]

Nice, I used the same initial inequality.

[@not_vinkami]

My way is a bit different As logarithm is strictly increasing, I take log[2] on both sides so I'm comparing 83 and 31×log[2](7). Since logarithm is strictly concave, we have log[2](7) > 2 + (7-4) / (8-4) = 2.75 (that is, making a straight line that passes through (4, 2) and (8, 3) and saying that log[2](x) is larger than that for 4 < x < 8) Now going back to the question, we have 31×log[2](7) > 31×2.75 = 85.25 > 83. Therefore, 7^31 > 2^83.

[@Hamm_y]

I used this approach which i think works well for a quick comparison. I quickly thought that 7 is roughly 2^2.8 ish since 2^2 is 4 and 2^3 is 9. 2.8*31 = 62+8*3.1 = 86.8, which is larger than 83. Thus 7^31 > 2^83. Even if you went lower by assuming 7 is roughly 2^2.7 (not 2.8), you still end up with 83.7, which is larger than 83.

[@ezequielgambaccini6754]

If you use logarithms in base 2, it’s much more simple, you end up with 83 on the one hand, and log2(7)*32 on the other, which leaves you with ~87, therefore proving it

[@RyanLewis-Johnson-wq6xs]

Input 7^31 = 157775382034845806615042743 Result True Logarithmic form 31 log(7, 7) = log(7, 157775382034845806615042743)

[@RyanLewis-Johnson-wq6xs]

Input (2^14)^6 = 19342813113834066795298816 Result True Logarithmic form 6 log(16384, 2^14) = log(16384, 19342813113834066795298816)

[@RyanLewis-Johnson-wq6xs]

Input (7^5)^6 = 22539340290692258087863249 Result True Left hand side (7^5)^6 = 22539340290692258087863249 Right hand side 22539340290692258087863249 = 22539340290692258087863249 Logarithmic form 6 log(16807, 7^5) = log(16807, 22539340290692258087863249)

[@radadadadee]

There's a lower exponent when they're almost equal: 2^3 ~ 7^1, I would've started there

[@sebastiencelma234]

2^84=2^3(28)=8^28 compare with 7^31 , divise all by 8^28 , compare 1 with ((7/8)^28)x343 , ((7/8)^28)x343>1 ; We can say without difficulties that 7^31>2^84>2^83

[@KaranTiwari-402]

why do you guys solve this easy ques with complex steps just put 7 as 2^3-1 and your que just solve in single step.👍

[@jackkalver4644]

It can even be shown that 7^31>2^85 (7^30>2^84)*(7>2)=(7^31>2^85)

[@Qermaq]

Hmm. Looking at powers of 7, we have 49, 343, 2401, 16807. 7^5 is 16807, and that's close to 2^14 = 16384. 7^5 > 2^14. If I multiply each exponent by the same number, the inequality should remain. So 7^30 = 7^(6*5) and 2^(6*14) is 2^84, so 7^31 > 7^30 > 2^84 > 2^83.

[@robertveith6383]

Did you write that without watching the video? You wrote essentially what is in the video.

[@Qermaq]

@@robertveith6383 I did, and it was uncanny that he did the same solution.

[@RyanLewis-Johnson-wq6xs]

I did it in my head.

[@RyanLewis-Johnson-wq6xs]

2^83

[@williamperez-hernandez3968]

Same method shows 7^31 > 2^86 (8x2^83). Since 7^30 > 2^84 and 7 > 4 (2^2), then 7^31 > 7 x 2^84 > 2^2 x 2^84 = 2^86. Thus, 7^31 > 2^86, or 8×2^83.

[@tejpalsingh366]

2^93/2^10 vs 7^ 31. Cears second is bigger

[@RyanLewis-Johnson-wq6xs]

2^14=16384

[@RyanLewis-Johnson-wq6xs]

7^5=16807

[@phill3986]

✌️😎😃👏👍😎😃✌️👏👍

[@SidneiMV]

7⁵ = 16807 2¹⁴ = 16384 7⁵ > 2¹⁴ => 7⁵/2¹⁴ > 1 2⁸³ = (2¹⁴)⁶2⁻¹ 7³¹ = (7⁵)⁶7 7³¹/2⁸³ = (7⁵/2¹⁴)⁶7¹2¹ > 1 *7³¹ > 2⁸³*