Only Geniuses Can Solve This | A Challenging Exponential Equation

[@divermike8943]

X=2 or X=1. By inspection. Can I get my genius certificate now?😂

[@miowacity]

I don't know why he is going so nuts with logs it says right there x=2 otherwise the x^2 exponent would be something different. In some math's 0 might work but usually 0^0 is considered 1

[@ericmiller6056]

@@miowacity Usually 0^0 is considered undefined.

[@ericmiller6056]

Yeh, exactly. I suppose we could have asked for a proof that those are all the solutions, but even that is obvious by geometric inspection.

[@miowacity]

@ericmiller6056 put it in your calculator and see what happens. Most likely it will say 1. This is the way it is in combinatorics as well because the empty set always has 1 element.

[@ericmiller6056]

@@miowacity You're confusing 0^0 and 0!. The latter is indeed defined and set equal to 1. If you put 0^0 into your calculator, the answer will be "error", just as it is if you ask it to divide by 0, and basically for the same reason. The empty set does not have one element; it has no elements; that is why it is the empty set. However, the set containing the empty set does indeed have one element, namely the empty set.

[@okaro6595]

Why did you distribute the ln x and then factor it out?

[@fredreisman8408]

April 1st is a ways away.

[@markhorrocks2531]

X=1

[@Bronzescorpion]

You don't strike me as a genius the way you complicate things. You even introduce problems. If x was 0 you would have divided by 0 in the first step. You should always avoid that. Why not take the logarithm from the start and get x*ln(x) = 2*ln(x)? From there you can easily see that either x has to be 2 for the two expression to be the same or ln(x) needs to be 0, therefore x would be 1.

[@lanecook7475]

X = 2