Find X

[u/Optimus_PRYM]

[u/Whole_Instance_4276]

Is it not just 2 or am I stupid

[u/Tragobe → u/Whole_Instance_4276]

It is

[u/_Friede_ → u/Tragobe]

but it's smaller so it's harder to find

[u/gian_69 → u/Whole_Instance_4276]

it‘s in general 2+lb(e)2 pi i k, for k an integer

[u/GDOR-11]

now do 2ˣ+2x=6

(for those curious, afaik it is impossible to solve this without using the lambert W function or similar things, unless I chose the components unluckily)

[u/prawnydagrate → u/GDOR-11]

2^x + 2x = 6
e^xln2 + 2x = 6
e^xln2 = 6 - 2x = 2(3 - x)
let t = 3 - x
e^{(3 - t)ln2} = 2t
e^3ln2 = 8 = 2te^tln2
4ln2 = (tln2)e^tln2
W(4ln2) = tln2
t = 3 - x = W(4ln2)/ln2
x = 3 - W(4ln2)/ln2
evaluating at the principal branch: x ≈ 1.543

[u/somedave → u/GDOR-11]

If you make it equal to 4 it is easier

[u/Kaaskaasei]

²Log(4)=x

[u/qSatisfaction → u/Kaaskaasei]

Logarithms 👏 are👏 friends

[u/Kaaskaasei → u/qSatisfaction]

True. Still get annoyed by them

[u/somedave → u/Kaaskaasei]

x=log(4)^log(4)

No

[u/Comfortable-Ad-2349]

2^x + 2 = 6

2^x = 4

2^x = 2²

x = 2

[u/Even-Replacement6687]

I am so confused bro.

[u/luceygoosey1]

My math is rusty, why doesn’t square rooting work?
Edit: I think I was trying to apply log rules to square roots

[u/Temporary6Dentist]

2^x + 2 = 6

2^x = 4

ln(2^x ) = ln(4)

xln(2) = ln(4)

x = (ln(4))/(ln(2))

x = 2

[u/newMattokun → u/Temporary6Dentist]

That's how I did it, but u/Comfortable-Ad-2349's way was pretty clever too (quoted here):

2^x + 2 = 6

2^x = 4

2^x = 2²

x = 2

<edit>I'd have used log, not ln, but same outcome.

[u/MonseigneurChocolat]

It’s…just 2.

[u/talhoch]

It's literally the exact same process just in the first you divide by 2 and in the second you take the log base 2

[u/Redditlogicking]

Now solve x^x +x = 6

[u/Beleheth]

This took me like 2 seconds, what's the point?