An Interesting Exponential Equation

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PLAYLISTS 🎵 :

[bartolhrg]

6:06 you could even use log[5/3] (log[3] (5))
log[B] meaning log with base B

[SidneiMV]

3ˣln5 = 5ˣln3
(5/3)ˣ = (ln5)/ln3 = log₃5
x(log₃5 - 1) = log₃(log₃5)

x = log₃(log₃5)/(log₃5 - 1)

[bkkboy-cmeb]

3^x·log5 = 5^x·log3
(3/5)^x = log3/log5
∴x =log(log3/log5)/log(3/5)

[gelbkehlchen]

Solution:
5^(3^x) = 3^(5^x) |ln() ⟹
3^x*ln(5) = 5^x*ln(3) |/[5^x*ln(5)] ⟹
(3/5)^x = ln(3)/ln(5) |ln() ⟹
x*ln(3/5) = ln[ln(3)/ln(5)] |/ln(3/5) ⟹
x = ln[ln(3)/ln(5)]/ln(3/5) ≈ 0, 7475

[almanduku]

I have a question: could the answer also be

+♾️
and
- ♾️

??

[vladimirkaplun]

Again 7 mins on the wrong rout and 1 min on the right one.

[dvd]

5^3^x=3^5^x
iif
3^x log5 = 5^x log3
iif
x log3 + log(log5) = x log5 + log(log3)
iif
x(log3 - log5) = log(log3) - log(log5)
iif
x = [log(log3) - log(log5)]/(log3 - log5) = log(log3/log5)/log(3/5) = log(log_5(3))/log(3/5) 🤔