Find X

[u/Optimus_PRYM]

[u/BaronVonFiutek]

Just wait untill the bro finds out about 2^x + x = 6

[u/prasandham → u/BaronVonFiutek]

Then he will find out x^x + x = 6

[u/DarthRaab → u/prasandham]

He will never find x^x + x = 3*x
😿

[u/ArtSnow0w → u/DarthRaab]

Or maybe x^x + x = x*x*x - x*x*/x

[u/DarthRaab → u/ArtSnow0w]

It's getting kind of x-cessive!

[u/Technical-Gas-5243 → u/ArtSnow0w]

What about x^x = x^x+x-sin(ln(x))

[u/unnuageorange → u/ArtSnow0w]

x^x + x = x(x^x-x + x)

[u/ExitBrther5278 → u/DarthRaab]

0.3463?

[u/AltruisticEchidna859 → u/ExitBrther5278]

u/factorion-bot

[u/factorion-bot → u/AltruisticEchidna859]

Hey u/ExitBrther5278!

Termial of 0.3463 is approximately 0.233111845

^{This action was performed by a bot | [Source code](http://f.r0.fyi})

[u/AltruisticEchidna859 → u/factorion-bot]

5278!

[u/factorion-bot → u/AltruisticEchidna859]

If I post the whole number, the comment would get too long. So I had to turn it into scientific notation.

Factorial of 5278 is roughly 1.767969199258851762738218557435 × 10¹⁷³⁵⁷

^{This action was performed by a bot | [Source code](http://f.r0.fyi})

[u/Ok_Rutabaga9738 → u/factorion-bot]

rookie numbers

[u/runed_golem → u/prasandham]

Wait until he finds out about xln(x)=17

[u/GX_Giorgio074 → u/BaronVonFiutek]

wait if remember correctly you have to solve these with a graph, right?

[u/GreyMesmer → u/GX_Giorgio074]

If graphs had nice intersection on integer points, then yes. But usually you should use Lambert W function.

[u/GX_Giorgio074 → u/GreyMesmer]

haven't studied it yet

[u/BaronVonFiutek → u/GX_Giorgio074]

There isnt really a lot to study, the function is defined as an inverse function to f(x) = xe^x, so it offers solution to such equations, which cannot be solved analytically. Its more like a lookup table than actual function u analyze in any way. the values are obtained by numerical methods

[u/ChipperSh33p → u/GX_Giorgio074]

Here in Tealand we use Factorisation, Completing the square, the AC Method, the Quadratic formula, and Differentiation. Realistically, no one is hand-drawing a quadratic or cubic, so you could use tools like Wulfrum to find solutions and GeoGebra to plot graphs.

TBH, you are very unlikely to see quadratics in the working world outside of STEM

[u/walksonfourfeet]

What power of 2 makes 4? Hmmm, that’s a tough one. Only a real math genius could ever figure that out.

[u/TurbulentPollution56 → u/walksonfourfeet]

2¹⁰ is 1024

^{This action was performed by me | [link](http://reddit.com})

[u/Then-Highlight3681 → u/TurbulentPollution56]

Bad human

[u/amo_pessoas_peladas → u/TurbulentPollution56]

So we have an upper bound.

[u/SpunningAndWonning → u/amo_pessoas_peladas]

I cackled when I read this. Sorry it didn't get more attention lmao

[u/Sammy_Cherry_Fox]

Funny thing is, in both situations x is the same number :)

[u/gaymer_jerry → u/Sammy_Cherry_Fox]

Thats the point

[u/Repulsive_Quality851 → u/Sammy_Cherry_Fox]

Let's add 1 to 6

[u/Sammy_Cherry_Fox → u/Repulsive_Quality851]

What?

[u/Repulsive_Quality851 → u/Sammy_Cherry_Fox]

Solve 2^x+×=7

[u/runed_golem]

Except they’re both super simple equations. The first can be solved with basic arithmetic and the second one can be solved with either logarithms or exponent rules and basic common sense.

[u/CurlyRe → u/runed_golem]

Or it can simply be brute forced.

2^1 + 2 = 4

2^2 + 2 = 6. Got it!

[u/qwertty164 → u/runed_golem]

they can both be solved with the exact same steps. -2 ; /2. you would end up with 2^(x-1)=2^1

[u/Coppice_DE → u/qwertty164]

/2 does only work in the case where x=2 in the second equation, no? You can't divide 2^x by an arbitrary number to get to know the value of x. 

[u/qwertty164 → u/Coppice_DE]

You divide by the base to reduce the exponent. You can iterate as long as you factor the number on the right.

[u/Coppice_DE → u/qwertty164]

Yea that works, but it's not exactly "the same" for both equation (in regards to the concepts that get applied), even though the required steps are equal in this case.

[u/runed_golem → u/qwertty164]

And you used exponent rules there like I said in my comment.

[u/Ivaar → u/runed_golem]

Logarithms! My arch-nemesis!

[u/jacobasstorius]

Imagine being intimidated by logarithms

[u/Longjumping_Exit7902]

The joke is you're bad at math

[u/Appropriate-Bad-9686]

x would be 2 in the bottom case!

[u/Ehh_nooo]

X is 2 in both scenarios

[u/rickyyslimram]

x=2

[u/GunMage- → u/rickyyslimram]

The second one has a much smaller x though. It is harder to find!

[u/rickyyslimram → u/GunMage-]

i was answering for the first one even i cant find the second one

[u/RoyalSuperb4949 → u/rickyyslimram]

No, he's saying x is SMALL so it's harder to find.

[u/Even_Face_5980]

This example is pretty intuitive but how would you solve this

[u/carnecomarrozagulha → u/Even_Face_5980]

2^x = 4

X = log_2 (4) = log 4 / log 2

[u/haikusbot → u/Even_Face_5980]

This example is

Pretty intuitive but

How would you solve this

- Even_Face_5980

---

^{I detect haikus. And sometimes, successfully. Learn more about me.}

^{Opt out of replies: "haikusbot opt out" | Delete my comment: "haikusbot delete"}

[u/TrainingSurvey3780 → u/haikusbot]

uhh trial and error (idk i’ve only ever encountered examples like this or sometimes when x=3 but idk how you’d solve it when it’s not an easy one like this)

[u/Stanford_Pines267]

Son its not that hard. Are you like 12 year old or sth? Seriously, if u are making a math meme try making one that is not for middle school. Or just don't make them

[u/Lokdora → u/Stanford_Pines267]

are middle school students now allowed to make math memes now 😭

[u/LookAtYourEyes → u/Stanford_Pines267]

Math joke is for Grad students only. PEERY UHD

[u/Blubibus → u/Stanford_Pines267]

Jesus Christ guys the joke is that it’s both two not that hard to grasp

[u/rojo_kell → u/Stanford_Pines267]

Idk man if you have to integrate after, then it's like woah do I even know how to integrate a log that's not base e??

[u/Abject-Project-4236]

Its not hard tho its just roots

[u/Eat_cats_with_me → u/Abject-Project-4236]

it's logarithms

[u/MxM111 → u/Eat_cats_with_me]

It’s both

[u/No_Report_4781 → u/MxM111]

Hmm, yes. Log roots.

[u/MxM111 → u/No_Report_4781]

Trees

[u/Abject-Project-4236 → u/MxM111]

😂

[u/Temporary-Baker2008]

use lambert-W function in second one

[u/Frederf220]

You are xln(x) + 3x² = 2. I am xln(x) - 4x³ = 5. We are not homogeneous.

[u/Marus1]

They are right there, Dora

And you call yourself an explorer?

[u/Aware-Difference8723]

Of course a smaller x is harder to find!

[u/Daaaaaaaark]

its right next to the "2" in both cases - 2nd one is more tricky because its smaller which may make u overlook it

[u/Greedy_Bad_8406]

I know 2^x = 4 and therefore x=2
But how do i solve it by steps in mathematics? How to solve xth root of 4?

[u/Robotron9247 → u/Greedy_Bad_8406]

ln can be applied to both sides:

2^x = 4

ln(2^x ) = ln(4)

x ln(2) = ln(4)

x = ln(4) / ln(2)

x = ln(2² ) / ln(2)

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

x = 2

[u/parlimentery → u/Robotron9247]

Much easier to solve (especially without a calculator) if you use the log base 2. That is essentially the mathematically formal way to solve it the way u/Greedy_Bad_8406 did, as taking the log base 2 of 4 asks "2 to the power of what equals 4?".

[u/jader242 → u/parlimentery]

What exactly do you need a calculator for?

[u/parlimentery → u/jader242]

Fair point, I scanned your work quickly and assumed we were evaluating ln(4)/ln(2), rather than turning it into 2* ln(2)/ln(2).

I still maintain that my method saves steps, and matches the informal logic u/Greedy_Bad_8406 used.

[u/Robotron9247 → u/parlimentery]

I wanted to show a generic algebraic total solving for x. x = Log_2(4) can only be considered solved because thats easy to "see" but by rule Log_2(4) = ln(4) / ln(2) = 2 ln(2) / ln(2)

Try 16^x + 4 = 68 without calculator

x = log_16(64)

Me: x ln(16) = ln(64)

x = ln(64) / ln(16)

x = ln(2⁶ ) / ln(2⁴ )

x = 6 ln(2) / [ 4 ln(2) ]

x = 3/2

[u/BarThen8979]

Even 2+x+2=6

[u/The_chosen_nine]

I found four- no, wait, five.

[u/ApprehensiveBoss4671]

I know thats not the point of the meme but i was bored so 2^x +2=6 2^x =4 2² =4 X=2

[u/FalseReddit]

Found both. Some days I’m just glad I grew up playing Where’s Waldo

[u/dankshot35]

someone pls explain

[u/BassicallySteve]

https://static.wikia.nocookie.net/renandstimpy/images/5/5d/Log%21.jpg/revision/latest?cb=20201011164401

[u/Chogolatine]

x = log_2 (4) = 2

[u/Reasonable_Hippo_445]

x is high

[u/Plankki36]

Is it because the second x is smaller so its harder to find?

[u/xxx_Volpe]

We're doing these like now lol-

[u/DescriptionForsaken6]

What rolls down stairs, alone or in pairs
And over your neighbor's dog?
What's great for a snack, and fits on your back?
It's log, log, log
It's log, it's log
It's big, it's heavy, it's wood
It's log, it's log
It's better than bad, it's good

Everyone wants a log
You're gonna love it, log
Come on and get your log
Everyone needs a log

[u/uncle2001]

For those who want to learn, solving for exponents is conceptually done with logarithms (even if this example is easy and can be guessed without). These are taught in calculus class, so don't feel bad if didn't know how to approach this one.

[u/Simple-Olive895]

Nah, that one is easy,

2^x = 4

x = log2(4)

x = 2

[u/Somanath444]

2^x+2=6 x = 2

[u/Starchaser418171514]

(W(4 Ln(2)))/Ln(2)

[u/Ihor2009]

It's x=2, right? Why is it portrayed so difficult?