I think this problem is not hard at all.

Here 95 % refers to kindergarten students 😂

I did this in my head.

Beautiful explanation. Well produced. Thank you.

*@ Math Window* ln(x) >= 0 ==> x >= e^0 ==> x >= 1 Your domain did not go far enough. This post expanded another post that mentioned your incorrect domain.

Yaayy!!! Solved it correctly!!!

By squaring and exponentiating results you are expanding your domain to solve a domain questionable change during the solution of the problem. In square root exoressions considered a positive solution by squaring a result introduces the negative root as an extraneous solution "liberal mathematicians" say is invalid. In Physics and Electrical Engineering mathematics "extraneous solutions" mathematics is an applied concept because of direction of vectors notations for vector math. Because a "-" result can happen with each "+" result in vector math where a square root was initially considered a positive, Physics of the sciences in mathematics define rules such as "vector right hand rule" to define the vector's direction even though the sauare root took a square root of a positive magnitude direction. Further academics of teachers explaining vector "right hand rule" for the vector "cross product" further answers this thought process more in the "mathenatical sciences!" ✓(lnx) = (1/2)(lnx) is (lnx) = 1/4((lnx)^2) by squaring the left hand side and the right hand side of the equation "mathematical sciences" routinely perform to contain an extraneous solution or root for further analysis. If lnx doesnt equal 0 such as in the ln(1) case then 1/4ln(x) = 1 or ln(x) = 4 or x = e^4 other solution. x = 1 or x = e^4 both solve the initial given equation. In sauaring and exponentiating we know x = 1 or x = e^4 and since 1 is not equal to e^4 we know x is one of those solutions or the other. Because both solutions silve the given equarion there were no extraneous solutions that didn't solve the given equation. The graph of the two curves would have intersected twice in all domain of x (the domain expanded by the exponentiated process).

Yes, in fact, it has two solutions: x=1 and x=e^4.

But t>0 !!! How you took t=0 as a solution ?!!!

√lnx = (1/2)lnx ln²x = 4lnx lnx(lnx - 4) = 0 lnx = 0 => *x = 1* lnx = 4 => *x = e⁴*

your domain is wrong, because you said x>0 but if x=0.5 => lnx=1

Correct solution (aside from the wrong domain as already pointed out in other comments), though sorry, but I'm sick of these clickbait video titles. 🤦‍♂️ Why not just name the mathematical problem itself in a creative way (e.g. "logs & roots in one equation")? And do you have any statistical sources for your "95%" claim? It might make some people feel better thinking they are part of the 5% after they got the correct solution, but not me. I solved it by myself as well before watching your solution, but that doesn't make me proud in this way, but I rather think this problem definitely ain't that hard as you claim. So I don't only think that these "95%" statistics don't exist, but even if they did, they would be wrong. Not meant to offend you, but I hope you get my point. Have a nice day. ✌️

This is basic high school math

I am in the 5% (I had just watched the answer after I solve it and the answer was good) And I solved in 10 secs

This was so useless, you should see ln(x) and automatically know that at x = 1 it's zero.....

Took me about 2 seconds. Not hard at all in my opinion.

Bud I solved it orally wdym 95%failed💀💀💀

ummmm x = 1?

10000

√[ln(x)]=ln(√x) Domain x>0 ln(x)=[ln(√x)]² Let y=√x, y²=x ln(y²)=[ln(y)]² 2[ln(y)]=[ln(y)]² [ln(y)]²-2[ln(y)]=0 Let z=ln(y) z²-2z=0 z(z-2)=0 z=0 ln(y)=0 y=e⁰ y=1 √x=1 x=1 ❤ z-2=0 z=2 ln(y)=2 y=e² √x=e² x=e⁴ ❤

√( ln (.x)) = (1/2) ln ( x) ln ( x). = 4 x = e ^4

👢