Let's Compare 2⁸³ and 7³¹

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[trumpetbob]

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.

[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.

[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.

[ezequielgambaccini]

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

[radadadadee]

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

[RyanLewis-Johnson-wqxs]

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)

[RyanLewis-Johnson-wqxs]

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

[RyanLewis-Johnson-wqxs]

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

[sebastiencelma]

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

[jackkalver]

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

[KaranTiwari-]

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.👍

[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.

[williamperez-hernandez]

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.

[Quest]

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

[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*