...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...
Thank you very much i have an exam tomorrow and hopefully I’ll do well
Good Luck!
Thank you so much 😊
Can you pls provide with some more questions of somewhat higher level
They were really easy
@Ojasvi; What topics are you looking for? Another log example is ua-cam.com/video/_5RALX7_Ch4/v-deo.html but I guess that is too easy for you.
@@MathsWithJay some questions including rational inequalities and modulus in log
@Ojasvi; Can you give examples? Here are some simple inequalities: ua-cam.com/play/PLgQUIweMg9eIC4MQjWwUvFiiaPfOFYI7_.html
...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...
Yes, it is essential to look back at the original equation when checking!
I don't quite understand how to factor a polynomial like 8x² - 10x - 3 in a fast way
See ua-cam.com/users/shortsFfD-bpOsmV4 or ua-cam.com/video/sMj1GAc3hAU/v-deo.html
thank you so much! this video was really helpful :)! hopefully i'll do well on my log test tmr ahaha ;;
Good luck for your test!
Thank you, this helped consolidate my knowledge.
Great to hear!
where does 11-6x go
At what time in the video?
thank you very much
@Rhys: Thank you!
nice man, i loved it
@Ayman Zubair: Thank you!
Can't we just get rid of log2 and then say (11-6x)=2(x-1)+3
If only...!
@@MathsWithJay only
Thank you!!
Thank you!
Thankyou so much!
Glad it helped!
ok
OK
It's very easy
@Ankur: Too easy?
@@MathsWithJay too easy still 3 years later