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[Daily puzzle] Simple algebra problem.
1148 sats \ 17 comments \ @south_korea_ln 28 Oct 2024 science
As yesterday's (hard) puzzle (#741908) hasn't been solved yet (new hint given in #743081), here is a comparatively simpler problem for today.
Solve for x:
\frac{27^x+343^x}{63^x+147^x} = \frac{37}{21}
Previous iteration: #741908 (unsolved)
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500 sats \ 6 replies \ @SimpleStacker 28 Oct 2024
\frac{27^x + 343^x}{63^x+147^x}
= \frac{27^x}{63^x + 147^x} + \frac{343^x}{63^x + 147^x}
= \frac{3^x 9^x}{3^x (21^x + 49^x)} + \frac{7^x 49^x}{7^x (9^x + 21^x)}
= \frac{9^x}{7^x (3^x + 7^x)} + \frac{49^x}{3^x (3^x + 7^x)}
= \frac{9^x 3^x + 49^x 7^x}{3^x 7^x (3^x + 7^x)}
= \frac{ 3^{3x} + 7^{3x} }{ 21^x (3^x + 7^x) }
= \frac{ (3^x + 7^x)(3^{2x} + 7^{2x} - 3^x7^x) }{ 21^x (3^x + 7^x) }
= \frac{ 3^{2x} + 7^{2x} - 3^x7^x }{ 21^x}
= \left( \frac{3}{7} \right)^{x} + \left(\frac{3}{7}\right)^{-x} - 1
\left( \frac{3}{7} \right)^{x} + \left(\frac{3}{7}\right)^{-x} = \frac{37}{21} +1 = \frac{3}{7} + \frac{7}{3}
So the answers are x=1 or x=-1
Phew, this took me a while. There goes my workday >:(
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132 sats \ 5 replies \ @SimpleStacker 28 Oct 2024
when my colleagues ask me why i'm not publishing more, I'll have to tell them some physicist from South Korea has been sending me math puzzles....
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42 sats \ 4 replies \ @south_korea_ln OP 29 Oct 2024
When my colleagues ask me why I'm not publishing more, I'll have to tell them I must keep competing minion researchers in check by keeping them busy with math puzzles. If no one publishes, we all win. No? :)
EDIT: as I wrote this, my phone buzzed to tell me I got paid a paper bonus by the university. They haven't noticed anything yet~~
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21 sats \ 1 reply \ @SimpleStacker 29 Oct 2024
Congrats! My Econ prof friend in Korea told me about publication bonuses. Wish we had those at my institution
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21 sats \ 0 replies \ @south_korea_ln OP 29 Oct 2024
They're nice for sure. But incentivises quantity over quality sometimes. Also profs arguing over who can be corresponding author just because of that.
Some unis also are getting absurd with the amounts. Heard of over 200k dollars for a paper in Nature or Science. More the case in China though, i think.
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0 sats \ 0 replies \ @Bell_curve 29 Oct 2024
Genius strategy
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0 sats \ 0 replies \ @nym 29 Oct 2024
That’s just as important!
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400 sats \ 1 reply \ @aliceandthewonderland 28 Oct 2024
I believe there's two solutions for this, 1 and -1
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0 sats \ 0 replies \ @WeAreAllSatoshi 28 Oct 2024
I concur, but I don’t know how to prove it
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113 sats \ 1 reply \ @Undisciplined 28 Oct 2024
It's pretty obviously 1.
Solution: plug in 1 to verify the answer.
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32 sats \ 0 replies \ @south_korea_ln OP 28 Oct 2024
Indeed. But it's not the only solution.
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78 sats \ 5 replies \ @0xbitcoiner 28 Oct 2024
1^x = \frac{37}{21}
log1(37/21)?
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0 sats \ 4 replies \ @south_korea_ln OP 28 Oct 2024
Not sure how you got to this expression...
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0 sats \ 3 replies \ @0xbitcoiner 28 Oct 2024
\frac{(27*343)^x}{(63*147)^x} = \frac{37}{21}
(\frac{27*343}{63*147})^x= \frac{37}{21}
1^x = \frac{37}{21}
I can't remember how to solve x. (log?)
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194 sats \ 2 replies \ @south_korea_ln OP 28 Oct 2024
a^x + b^x \neq (ab)^x
You can easily check this inequality using a=2, b=3 and x=2.
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87 sats \ 1 reply \ @0xbitcoiner 28 Oct 2024
I've been looking at it wrong. Where do you have +, I saw x
a^x × b^x = (ab)^x
I need to replace the lenses in my glasses
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33 sats \ 0 replies \ @south_korea_ln OP 28 Oct 2024
Honest mistake ;)
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