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....


$$
\frac{27^x + 343^x}{63^x+147^x}
$$

$$
= \frac{27^x}{63^x + 147^x} + \frac{343^x}{63^x + 147^x}
$$

$$
= \frac{3^x 9^x}{3^x (21^x + 49^x)} + \frac{7^x 49^x}{7^x (9^x + 21^x)}
$$ 

$$
= \frac{9^x}{7^x (3^x + 7^x)} + \frac{49^x}{3^x (3^x + 7^x)}
$$

$$
= \frac{9^x 3^x + 49^x 7^x}{3^x 7^x (3^x + 7^x)}
$$

$$
= \frac{ 3^{3x} + 7^{3x} }{ 21^x (3^x + 7^x) }
$$

$$
= \frac{ (3^x + 7^x)(3^{2x} + 7^{2x} - 3^x7^x) }{ 21^x (3^x + 7^x) }
$$

$$
= \frac{ 3^{2x} + 7^{2x} - 3^x7^x }{ 21^x}
$$

$$
= \left( \frac{3}{7} \right)^{x} + \left(\frac{3}{7}\right)^{-x} - 1
$$

$$
\left( \frac{3}{7} \right)^{x} + \left(\frac{3}{7}\right)^{-x} = \frac{37}{21} +1 = \frac{3}{7} + \frac{7}{3}
$$

So the answers are $$x=1$$ or $$x=-1$$

Phew, this took me a while.  There goes my workday >:(

SimpleStacker

I concur, but I don’t know how to prove it 

I believe there's two solutions for this, 1 and -1

Alice_and_the_Wonderland

Not sure how you got to this expression...

0xbitcoiner

It's pretty obviously 1.

Solution: plug in 1 to verify the answer.

Undisciplined

) hasn't been solved yet (new hint given in 

), here is a comparatively simpler problem for today.

science

As yesterday's (hard) puzzle (https://stacker.news/items/741908/r/south_korea_ln) hasn't been solved yet (new hint given in https://stacker.news/items/741908/r/south_korea_ln?commentId=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: https://stacker.news/items/741908/r/south_korea_ln (unsolved)