Thank you so much 😊 Can you pls provide with some more questions of somewhat higher level They were really easy
@MathsWithJay
5 жыл бұрын
@Ojasvi; What topics are you looking for? Another log example is kzitem.info/news/bejne/wGuIp4COcJF0m2k but I guess that is too easy for you.
@ojasviagrawal3180
5 жыл бұрын
@@MathsWithJay some questions including rational inequalities and modulus in log
@MathsWithJay
5 жыл бұрын
@Ojasvi; Can you give examples? Here are some simple inequalities: kzitem.info/door/PLgQUIweMg9eIC4MQjWwUvFiiaPfOFYI7_
@blackchicken2243
2 жыл бұрын
Thank you very much i have an exam tomorrow and hopefully I’ll do well
@MathsWithJay
2 жыл бұрын
Good Luck!
@alastairhart5308
Жыл бұрын
Thank you, this helped consolidate my knowledge.
@MathsWithJay
Жыл бұрын
Great to hear!
@waffles1042
4 жыл бұрын
nice man, i loved it
@MathsWithJay
4 жыл бұрын
@Ayman Zubair: Thank you!
@d1nyamic
Жыл бұрын
thank you so much! this video was really helpful :)! hopefully i'll do well on my log test tmr ahaha ;;
@MathsWithJay
Жыл бұрын
Good luck for your test!
@astron-out4982
6 жыл бұрын
Thank you!!
@MathsWithJay
6 жыл бұрын
Thank you!
@c6ldass781
Жыл бұрын
I don't quite understand how to factor a polynomial like 8x² - 10x - 3 in a fast way
@MathsWithJay
Жыл бұрын
See kzitem.infoFfD-bpOsmV4 or kzitem.info/news/bejne/1IOgl3t3nGWZdIo
@RhysTucker2603
5 жыл бұрын
thank you very much
@MathsWithJay
5 жыл бұрын
@Rhys: Thank you!
@MelOkunbor
Жыл бұрын
Thankyou so much!
@MathsWithJay
Жыл бұрын
Glad it helped!
@jan-willemreens9010
2 жыл бұрын
...Good day Miss Jay, After a thorough examination of the second problem, I would like to make a comment that it is INDEED important to look at the ORIGINAL EQUATION: log(2)(11-6x)=2log(2)(x-1)+3 for the determination of the restrictions for x: 11-6x>0 and x-1>0, with which finally x=3/2 is the only solution, and x=-1/4 becomes invalid. However, if instead of the original equation you now look at the algebraically manipulated equation: log(2)(11-6x)=log(2)[(x-1)^2]+3, then the restriction x-1>0 expires, with the result that next to x=3/2, x=-1/4 also becomes a valid solution! After all, the graph of y1=2log(2)(x-1)+3 is different from the graph of y2=log(2)[(x-1)^2]+3, while you actually only have used one of the valid log properties (laws) for logarithms: 2log(2)(x-1)+3=log(2)[(x-1)^2]+3. But in the end that makes a big difference in the outcome of the equation! So, one has to be very careful when setting the restrictions for x, and therefore only look back to the original equation, Miss Jay. To be sure, I also looked at both options graphically: 1) (the original situation) y3=log(2)(11-6x) with y1=2log(2)(x-1)+3 ---> (one intersection x=3/2), and 2) y3=log(2)(11-6x) with y2=log(2)[(x-1)^2]+3 ---> (two intersections x=3/2 and -1/4). Thank you for your (not too difficult?) presentation, Take good care, Jan-W p.s. At least one has to be aware of using the right restrictions, and therefore only looking at the original equation...
@MathsWithJay
2 жыл бұрын
Yes, it is essential to look back at the original equation when checking!
@gabrielkendall4349
Жыл бұрын
where does 11-6x go
@MathsWithJay
Жыл бұрын
At what time in the video?
@cutegirl-cb9ol
Жыл бұрын
Can't we just get rid of log2 and then say (11-6x)=2(x-1)+3
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