By the definition of logarithm, it is directly 2^2 = x^2/(x+3); then this becomes quadratics equation. It ends up with x = 6 and x =-2; You are a great teacher. I wish I had the teacher like you in my high school, explaining everything clearly. Cheers!
great explanation but for some reason im having some trouble with a similar equation where you add the 2 in front of the log aside from this one that multiplies. i have checked everywhere and cant find a clear source. thank you though
Can someone help? Why did you reject the solution x=-2? I know logs can’t take negative inputs, however that 2 out front of the original equation means we would take x^2, therefore the input would always be positive. That’s why I thought it would still be a solution.
Also, at 6:16 you said that log2 (6) works, but log2 (9) doesn’t work... why doesn’t that work? Assuming it doesn’t work, then why didn’t you reject the solution x=6?
Shouldn't both be correct? logBase2 of x^2 where x=-2 is logBase2 of 4, minus logBase2(x+3) (which works out to be zero) is equal to 2. For 6, 2logBase2(6)-logBase2(9) will equal two as well.
@@brianmclogan 4 days away from 9th anniversary, going through most of your log videos for a test tomorrow, you are an absolute lifesaver for putting these on youtube. My math teacher's explanations make me want to blow my brains out. I wish I was in your class, I'd probably get an A+.
Brian McLogan first of all it turned out that there are two solutions : x =6 and x = -2 ... Near the end of the video you said that if we substituted x = 6 in Log base 2 of ( x + 3 ) it will give us Log base 2 of 9 and it won't work but that's incorrect because Log base 2 of 9 = 3.1699 . Secondly , what I want to understand : why is x = -2 not accepted since 2Log base 2 of x = Log base 2 of (x)^2 = Log base 2 of (-2)^2 = Log base 2 of 4 =2 so why is it unaccepted ?
thank you for being straightforward, long lectures on math just always goes in one ear and out the other so things like this help a lot
God bless!!
By the definition of logarithm, it is directly 2^2 = x^2/(x+3); then this becomes quadratics equation. It ends up with x = 6 and x =-2; You are a great teacher. I wish I had the teacher like you in my high school, explaining everything clearly. Cheers!
God bless! B
youve helped me through my 10-week online college math course. thank you 😭🙏 im rewatching these before the final tomorrow
God bless !!
Headphone users = death by keys
it was so LOUD
"Kiss" of death
Very Good, Brian McLogan, Keep It Up! 🎉🎉🎉🎉
great explanation but for some reason im having some trouble with a similar equation where you add the 2 in front of the log aside from this one that multiplies. i have checked everywhere and cant find a clear source. thank you though
happy to help
you are the BEST! thank you so much for your videos!!
Goodmorning sir, could you please make a video about on how to solve logarithmic inequalities? Thank you
The videos from 2006 are helping
Thank you so much
You are amazing.
Hope you were my teacher.
Thank you happy to help
log5+log(2x+10)-2=log(x-4)
you have saved me brian. god bless.
Can someone help? Why did you reject the solution x=-2? I know logs can’t take negative inputs, however that 2 out front of the original equation means we would take x^2, therefore the input would always be positive. That’s why I thought it would still be a solution.
Also, at 6:16 you said that log2 (6) works, but log2 (9) doesn’t work... why doesn’t that work? Assuming it doesn’t work, then why didn’t you reject the solution x=6?
@@crod1232 , he misspoke.
ruclips.net/video/0K5p49E7TWg/видео.html
@@jdilksjr so when he misspoke, he meant it X=6 does work?
@@crod1232 That will work, he misspoke. You can use the rule and see the negative sign as (x+3)^-1, which I believe will give you the same answer
thank you so much man
Shouldn't both be correct? logBase2 of x^2 where x=-2 is logBase2 of 4, minus logBase2(x+3) (which works out to be zero) is equal to 2. For 6, 2logBase2(6)-logBase2(9) will equal two as well.
you can’t evaluate 2log(-2). you have to plug your x value into the original equation
goog video,thanks
Got like 25min before my final 😂
I’m doing a question where log2(x+5/x)=3 how would I get my answer from that
Thank you for your tutorial in log but it is not visible at all
Thank U ♥♥♥♥
you are very welcome!
هلا وغلا
Happy anniversary of this video
cheers!
@@brianmclogan 4 days away from 9th anniversary, going through most of your log videos for a test tomorrow, you are an absolute lifesaver for putting these on youtube. My math teacher's explanations make me want to blow my brains out. I wish I was in your class, I'd probably get an A+.
Long 7-Minute Video
what did you do with the two in log x. Where did it go?
which one? at what point?
Why is 6 still an answer if 6 and 3 =9 and that doesn't work?
Dude -2 in the equation is going to be squared and become positive
what time point?
Brian McLogan first of all it turned out that there are two solutions : x =6 and x = -2 ... Near the end of the video you said that if we substituted x = 6 in Log base 2 of ( x + 3 ) it will give us Log base 2 of 9 and it won't work but that's incorrect because Log base 2 of 9 = 3.1699 . Secondly , what I want to understand : why is x = -2 not accepted since 2Log base 2 of x = Log base 2 of (x)^2 = Log base 2 of (-2)^2 = Log base 2 of 4 =2 so why is it unaccepted ?
@@mohammedkhaled2456 because it will become and imaginary number
@@decoyreggieyt4621 I think they mean if you put the 2 back in so it becomes log2(-2^2)
Not visible please.
What?
Why are they torturing us?
i think i have adhd
why don't you ever do the problems I have on my actual homework? I have a equal sign in the middle, separating two logs.
Thank you so much