Log equations worked exam question

Edexcel January 2010 C2 Question 5, including checking answers (part (a) on a Casio calculator)

[ojasviagrawal]

Thank you so much 😊
Can you pls provide with some more questions of somewhat higher level
They were really easy

[blackchicken]

Thank you very much i have an exam tomorrow and hopefully I’ll do well

[alastairhart]

Thank you, this helped consolidate my knowledge.

[jan-willemreens]

...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: 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...

[dnyamic]

thank you so much! this video was really helpful :)! hopefully i'll do well on my log test tmr ahaha ;;

[cldass]

I don't quite understand how to factor a polynomial like 8x² - 10x - 3 in a fast way

[cutegirl-cbol]

Can't we just get rid of log2 and then say (11-6x)=2(x-1)+3