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Thank you! ❤❤
You are welcome sir
Chiwendu, abeg, that's alright. Too much for that small thing.
🤣 okay sir.
1^x=2 x=i(Ln(2))/(2πn)= iLog[e^(2πn),2] It’s in my head.
👍👍
Logarithm is the way to go
cant divide by 0
Yeah
ln(1)/ln(1)=0/0
👍🙏🙏
Sorry log of any base of 1 is equal 0
Thank you
@@MathswithChinwendu You are welcome !. I think to use Complex Roots where z^x=w as example. Let me look for a solution later on.
Sorry. Dead wrong.
👍
What're you waiting for? By now you should have became a rocket scientist😬🙄🙄🥱
🤣
Thank you! ❤❤
You are welcome sir
Chiwendu, abeg, that's alright. Too much for that small thing.
🤣 okay sir.
1^x=2 x=i(Ln(2))/(2πn)= iLog[e^(2πn),2] It’s in my head.
👍👍
Logarithm is the way to go
👍👍
cant divide by 0
Yeah
ln(1)/ln(1)=0/0
👍🙏🙏
Sorry log of any base of 1 is equal 0
Thank you
@@MathswithChinwendu You are welcome !. I think to use Complex Roots where z^x=w as example. Let me look for a solution later on.
Sorry. Dead wrong.
👍
What're you waiting for? By now you should have became a rocket scientist😬🙄🙄🥱
🤣