Solving 2^x=x+(1/x), a Non-Standard Equation

If you need to post a picture of your solution or idea:
#ChallengingMathProblems #NonStandardEquations

PLAYLISTS 🎵 :

[johnlashlee]

2^x = x + 1/x
2^(1/x) = 1/x + 1/(1/x) = x + 1/x
Combine them: 2^x = 2^(1/x).
So, x - 1/x = 0 --> x either -1 or 1. Checking the solutions eliminates the x = -1 case.
So x = 1.

[moeberry]

Another way to do it Syber without using calculus is to graph 1/x+ x by using strictly vertical and horizontal asymptotes but since the degree of the numerator is larger by 1 than the denominator we actually have a slanted asymptote at y=x. And after finding the y intercept it’s very easy to sketch this graph and see that there is only 1 solution.

[WahranRai]

Expression gives : 2^x - 1/x = x ---> find fixed point of f(x) = 2^x - 1/x (fixed point means f(x) = x ) ---->
derivative(f(x)) = ln2exp(xln2) + 1/x^2 > 0, if you plot f(x) you will see that :
x < 0 : f(x) > 0 and no intersection with y=x (first bisector)
x> 0 : f(x) is strictly increasing and intersects with y=x in x=1 : y=1 and f(1) = 2^1 - 1/1= 1 ----> *x=1 unique solution*

[threstytorres]

2^x is a natural number, x + 1/x must be a natural number as well, in order to x + 1/x to be a natural number, 1 needs to divides x ( x|1 is a natural number ), and therefore x=1

If x=1 we have;
2¹ = 1 + 1/1
2 = 1 + 1
2 = 2

Therefore 1 is the ONLY solution

[snejpu]

I managed to solve this problem using very weird method involving inequalities. For x<0, x + 1/x <= -2 (trivial proof) and 2^x is always >0, so no solutions for x<0. For x=0 the equation is undefined. For x>0, x + 1/x >= 2 (trivial proof), for x from 0 to 1 it decreases and for 1 to infinity it increases. On the other hand, 2^x is monotonic and increases to meet the other function for x=1 (trivial solution). To see that for x>1, 2^x increases faster than x+1/x, use derivatives. 1-1/x^2 is less than ln(2)*2^x, because for basic case of x=1 it is already clear and it will proceed, because 2ln(2) is already greater than 1, which is the maximum value of 1-1/x^2.

[WaterMeetsLava]

This was a pretty easy equation, as if you graph the two functions, there is only one solution. Which means an appropriate guess-and-check method would work. And so x=1 is the only solution.

[frentz]

"And how do we go about solving these kind of things? I have no idea. My method was to make the numbers really nice in the original problem and then guess the answer."

[flaviuzzz]

This was great I've learned about asimptotes a while back and i was happy to see them again

[MrLidless]

There are an infinite of complex answers as well of the form x = a ± bi, where:

a^2-b^2 +1 and
2ab

Note that (a, b) = (1, 0) is a solution as per the video.
To six decimal places, another solution is 0.344128 ± 1.738946i.

[carloshuertas]

Hey SyberMath! I looked at this math problem and immediately guessed that x=1. Thanks a lot!

[seegeeaye]

We just need to prove the uniqueness of the solution x = 1.. We can prove it by comparing the values of both sides. First we know both sides as graphs, they intersect only in the first quadrant, therefore we just need to compare when x > 0. LHS < RHS when x < 1; LHS > RHS when x > 1, ie. Unique solution x = 1

[tayserbinjafor]

Can't you use the Lambert W function?
Here it is:
W(x × e^x) = x
W(x) × e^W(x) = x
Here, W(x) is the Lambert W function of the variable 'x'.

[sukantasarangi]

Hey syber,
I have done another method to find out the answer..
Here is how it goes..
First i took the eqn:
2^x = x +1/x
Dividing both sides by x, we get
2^x/x = 1+1/x²
Rearranging terms, we get:
2^x/x - 1/x² = 1
Taking the LCM
[(2^x)^x - 1]/x² = 1
(2^x)^x - 1 = x²

Now taking (2^x)^x -1 as 2^x² -1
Now drawing the graph by takin eqn..
2^x² - 1 =y
And
x² = y
I get two values from the graph. i.e x = 0 and x = 1
But x = 0 wont satisfy as the answer would be undefined..
Hence x = 1
Thnx syber for giving these probs..
Lots of love

[robertlunderwood]

Before putting pencil to paper, my first thought was that x + 1/x would be act linear-ish for x < -1 and x > 1, so my main concern would be in between, especially with that asymptote. I figured that if I could graph that out, I could prove that there is only the one intercept.

I was right.

[lorenzopasini]

I would have solved it firstly noticing that x has to be greater than zero, then I have 2^x= x+1/x and even 2^1/x=x+1/x, so I have that 2^x=x+1/x=2^1/x, that implies 2^x=2^1/x, which implies x=1/x with x>0, so x^2=1 with x>0, so x=1

[NathanSimonGottemer]

Since the right side asymptotically approaches the y=x line, and y=2^x is so much greater than y=x for x>1, it follows that since there is no solution lower than 1, x=1 should be the only real solution

[holyshit]

x = 1 is the obvious solution, easy to guess
but maybe there are other solutions with Lambert W function

[chocolatecandycake]

This is a complex equation. It is non-standard, as in it is impossible to calculate an answer based on the data given. What we can is guess or do check and trial and error.

For example, your equation is 2^x = x + 1/x. The answer is 1 only because it is a coincidence. But if we change it to 1.9^x = x + 1/x it is virtually inpossible to pinpoint an answer.

[frentz]

"We can't solve this problem by normal means, we're gonna have to .. " guess the answer! guess the answer! lol

[fatihsrk]

Wish you redo the 2^x=x^2 video with the negative solution’s addition