Two Problems In One | Math Is Fun 😉

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[@farhansadik5423]

Man, you've become so much more expressive now and it's so fun listening to you talk. Also love the "don't ask y" or "2b or not 2b" jokes. You've made me love math at such a intricate level too. I just got my results for my state exam. I got a good result! So, thank you.

[@mtaur4113]

I always use parentheses for log(x^n). Even though the convention of unwritten parentheses in this case is fairly strong, I have seen enough calc 1 students fumble even when the parentheses were written out that I don't trust it to just be understood as another OOO thing.

[@scottleung9587]

Wow, that was impressive!

[@rickdesper]

Let y = log x. Solve y^n = n*y. Then x = e^y.

[@robertveith6383]

@ SyberMath -- The logarith is a function, so you should put the arguments inside of grouping symbols: [log(x)]^n = log(x^n).

[@SyberMath]

I just don't like it and I'm hoping it's understood but you're right!

[@BorisRaifler]

It seems to me, that it should be a maximum value for X when we try to solve for n

[@quelquun1366]

Can't we make -ln(a)/a = -ln(a) * e^(-ln(a)) ? to get that one of the solutions is always 1 (+ the other solution found with the other branch of the function).

[@stephenshefsky5201]

Explicitly, n = -W(-ln(a)/a)/ln(a) = -W(-ln(a)*e^(-ln(a)))/ln(a) = -(-ln(a))/ln(a) = 1. That's the main branch.

[@Rickest_Wojt]

What is why 😂

[@vladimirkaplun5774]

How to solve x=1 using Lambert 🤑

[@gauravroy8528]

One correct answer, x = 1 Explanation:- log 1 = 0 (log 1)^n = 0^n = 0 1^n = 1 log (1^n) = log 1=0 Hence x = 1 Have a nice day.