Solving x^{1/ln(x)}=e^{x-1}

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PLAYLISTS 🎵 :

[moeberry]

The domain is not just positive x values but also x cannot equal 1. Also the graph that you showed should have a hole at x=1. The coordinate is (1, e) where the hole occurs.

[Psykolord]

I wound up jumping straight to method 2 when working on it. I was actually a bit leery at first since it's been so long since I worked with logarithms, and was going "There's...there's no way it's that clear-cut, right?"

Lo and behold, it was.

[pyctyurypupiline]

My first instinct was to change base:
1/ln(x) = ln(e)/ln(x) = log(e) in the x base.

After that the answer just appears.

[envjvru]

x=exp(lnx) so x^(1/lnx) is just exp(lnx/lnx)=e which is the base of ln. So a nice logarithm property we get from this is
x^(1/logx) = b for all x>0
And where b is the base of the log

[Exonorm]

x^[1/lnx] = e for {x > 0, x =/= 1}, and since from the right side x=2 is the only value of x that will also make equal to e, then x must be 2

The reason this works is because 1=lne. Replacing that means you can use log properties on the left side exponent to create x^[log_x(e)], of course, the exponential and logarithm cancel leaving just e. The caveat is that the base of a logarithm must be greater than 0 and not equal to 1, which is another explanation for where the restrictions come from.

[hustyfoxx]

1/ln x = ln e / ln x
ln e / ln x = ln x e { log with base x }
Therefore x^(1/lnx)= x^(ln x e) = e
This explains why it was a constant.

[rafaeldiazsanchez]

You create lovely videos to review long forgotten High School maths. Thanks a lot.

[rajeshbuya]

OH MY GOD!!! I know 2nd method is always shorter, but never expected it to be just a 2-step method. How did I NOT think of it?

[stevenlitvintchouk]

I used the second method. Because of the ln, the domain must be x > 0. The left-hand side reduces to a constant e, and the derivative of the right-hand side must be positive. So an expression consisting of the right-hand side minus the left-hand side must always increase for X>0. Therefore, once we find the solution x = 2, there are no other solutions where the expression crosses the X-axis again.

[noahvale]

Another very nice video. Like it when you include the graphs.

[sohelzibara]

Here comes a third approach. To begin with and in order for the equation to defined, x has to be strictly greater than zero as well as different from 1. With that being said, the LHS is by definition equal to e and the equation turns into e = e^(x – 1), that is 1 = x – 1, that is x = 2.

[blaz]

I had no idea why this video was so long, because I immediately jumped to the second method. I thought that I must have been making some massive mistake, but nope. Two methods, one far slower than the one I found.

[michachyra]

Log a b times log b a =1, hence lhs=e, so x-1=1, x=2

[troys]

Not entirely sure but by raising both sides by ln(x), getting the solution of x = 1 would be simply raising both sides by 0 or something in those lines

[natashok]

Thank you. Very important to do the control of the Every equation! 🐶

[upulwijesingha]

I used the second method and just did it in mind !!😏

[tbg-brawlstars]

By a look, I solved that 2 is definitely a solution.

[DrQuatsch]

I rewrote the 1/ln(x) as ln(e)/ln(x). This is a bit iffy, but this can be seen as a base change fraction for log_x (e), the logarithm with base x. x^log_x(e) = e, with x>0 and x can't be 1. So then you just get e = e^(x - 1), which means x-1 = 1 and therefore x = 2.

[mcwulf]

I just jumped in with ln() each side which is simply x=2.

[threstytorres]

x^(1/Lnx) = e^(x-1)

Take Ln Both Sides then Simplify to get the answer
Ln[x^(1/Lnx)]= Ln[e^(x-1)]
=> 1/Lnx * Ln(x) = x-1
=> 1 = x-1 (The 1 in the Lhs comes from multiplyinh Lnx from its reciprocal)
=> x= 2 (add 1 to both sides)

Thus 2 is the Solution