in school I still haven't done logarythms in depth yet, but I kinda knew what they did, and as soon as you revealed that loga(x) - loga(y) = loga(x/y) I quickly found the answer. Thank you for teaching me a property of logarithms! :)
@@anglaismoyen still haven't done logs in school, I do them next year, but I remember watching 1 or 2 videos about logs a few months ago and had forgot most of the properties. the ones I still remember are: log(a^b) = blog(a) loga(a) = 1 e^ln(x) = x and then the ones I re-learned now. I'm still 15 and I honestly can't wait to get into this subject :)
You can use L'hopital's rule in so so many calculus questions, but i love how this guy tries to solve questions by other methods unless nothing else works, because its rather boring to use L'hopital's rule
@@itachu.you’re treating log_3 as the natural log, so those would not be the derivative, you’d need to apply chain rule and the derivative of log_a(x)=1/(x*log_e(a)) I’ll leave finding that derivative to you, if you feel so inclined. there’s an easier way to apply l’hopitals rule Lim (Log(3x+2)-log(x+2))=lim log((3x+2)/(x+2))= log( lim (3x+2 )/(x+2)) From here apply l’hopitals rule on the inside of the logarithm which Brings us to log_3 (lim 3/1)=log_3(3) =1
@@itachu. There has to be an infinity over infinity or a 0/0 for l'hopital to be applicable. In this case we have infinity minus infinity, which is not a condition that l'hopital's rule can be applied, so we have to perform some algebraic manipulation. If we use just the law of logs we find log(3x+2)-log(x+2)=log((3x+2)/(x+2)) so if we just focus on the value the inside is approaching to plug it into the logarithm we have infinity over infnity, meaning we are allowed to use l'hopitals rule so we have log_3(3)=1
in school I still haven't done logarythms in depth yet, but I kinda knew what they did, and as soon as you revealed that
loga(x) - loga(y) = loga(x/y) I quickly found the answer.
Thank you for teaching me a property of logarithms! :)
That rule is normally one of the first thing you learn about logarithms. How many lessons have you done on them?
@@anglaismoyen still haven't done logs in school, I do them next year, but I remember watching 1 or 2 videos about logs a few months ago and had forgot most of the properties. the ones I still remember are:
log(a^b) = blog(a)
loga(a) = 1
e^ln(x) = x
and then the ones I re-learned now. I'm still 15 and I honestly can't wait to get into this subject :)
yes
Excellent video! 😊
❤ Thank you sir, I love your content
Love your videos
we can use L' Hopital's rule.
You can use L'hopital's rule in so so many calculus questions, but i love how this guy tries to solve questions by other methods unless nothing else works, because its rather boring to use L'hopital's rule
how help
1/(3x +2) - 1/(x+2)
=> answers cleary zero help
@@itachu.you’re treating log_3 as the natural log, so those would not be the derivative, you’d need to apply chain rule and the derivative of log_a(x)=1/(x*log_e(a))
I’ll leave finding that derivative to you, if you feel so inclined. there’s an easier way to apply l’hopitals rule
Lim (Log(3x+2)-log(x+2))=lim log((3x+2)/(x+2))= log( lim (3x+2 )/(x+2))
From here apply l’hopitals rule on the inside of the logarithm which
Brings us to log_3 (lim 3/1)=log_3(3) =1
@@JadeNichols5270 hi
then I get
1/3(3x+2) - 1/(x+2)
after substitution of infinity I still get zero what difference does chain rule make
@@itachu. There has to be an infinity over infinity or a 0/0 for l'hopital to be applicable. In this case we have infinity minus infinity, which is not a condition that l'hopital's rule can be applied, so we have to perform some algebraic manipulation. If we use just the law of logs we find log(3x+2)-log(x+2)=log((3x+2)/(x+2)) so if we just focus on the value the inside is approaching to plug it into the logarithm we have infinity over infnity, meaning we are allowed to use l'hopitals rule so we have log_3(3)=1
Nice video as always!
never stop learning =)
How i ms this man so clever
With property of logatitm it becomes 1
1
🇮🇹
Harika