The Student Room Group

Core 2 Log question

Could anyone help me out in the question below, don't know where I have gone wrong: it's part b

ImageUploadedByStudent Room1399649303.175034.jpg

Here is my solution:

ImageUploadedByStudent Room1399649335.031301.jpg

Thanks :smile:


Posted from TSR Mobile
You were doing it right till you decided to factor out the 2 for the x ad 1 for some reason :tongue:. you should have added the 1 to the other side and then divided it all by 2.

So it should be ((ln.05/ln3) + 1)/2

Posted from TSR Mobile
(edited 9 years ago)
Original post by Numan786
You were doing it right till you decided to factor out the 2 for the x ad 1 for some reason :tongue:. you should have added the 1 to the other side and then divided it all by 2.

So it should be ((ln.05/ln3) + 1)/2

Posted from TSR Mobile


So do you always get rid of everything else, and then what x is divided by you take to the other side to find x?


Posted from TSR Mobile
Reply 3
Avatar for Sataris
Sataris
21
Original post by Jimmy20002012
So do you always get rid of everything else, and then what x is divided by you take to the other side to find x?


Posted from TSR Mobile


Well you don't have to, but when you divide 2x−12x-1 by 2 you get x−12x - \frac{1}{2} You divide every term by 2.
Original post by Jimmy20002012
So do you always get rid of everything else, and then what x is divided by you take to the other side to find x?


Posted from TSR Mobile


Can you solve 2x-1=5
Original post by TenOfThem
Can you solve 2x-1=5


2x= 6
X=6/2= 3



Posted from TSR Mobile
Original post by Jimmy20002012
2x= 6
X=6/2= 3



So yours was 2x-1 = something

SO you still have to +1 and then /2

:smile:
Original post by TenOfThem
So yours was 2x-1 = something

SO you still have to +1 and then /2

:smile:


Thanks for the help :smile:


Posted from TSR Mobile
Original post by TenOfThem
So yours was 2x-1 = something

SO you still have to +1 and then /2

:smile:


Could you also help me in this sequences question:
ImageUploadedByStudent Room1399660410.518533.jpg
ImageUploadedByStudent Room1399660422.157947.jpg

Surely if the 21st term is twice greater than the 16th terms, the equation applied to this question would be 2 x the 21st term??


Posted from TSR Mobile
Original post by Jimmy20002012
Could you also help me in this sequences question:
ImageUploadedByStudent Room1399660410.518533.jpg
ImageUploadedByStudent Room1399660422.157947.jpg

Surely if the 21st term is twice greater than the 16th terms, the equation applied to this question would be 2 x the 21st term??



Why do you think that?

a is twice b means a=2b
ImageUploadedByStudent Room1399742532.481914.jpg


Posted from TSR Mobile
If you had a question like

2 x 3^n-1 > 4 x 10^15 and you had to solve for n,

Would you apply logs and get

(log4 + log10^15 / log2 + log3) +1 (Log base 10 used)

Is this right or another way of doing it?? Thanks


Posted from TSR Mobile
Original post by Jimmy20002012
...


Method looks fine but see a mistake.

:smile:
Original post by lazy_fish
Method looks fine but see a mistake.

:smile:


What's the mistake??


Posted from TSR Mobile
Original post by Jimmy20002012
...


Just to clarify, your answer was n>log⁡4+log⁡1015log⁡2+log⁡3+1 \displaystyle n > \frac {\log 4 + \log 10^{15}} {\log 2 + \log 3} +1

If so, there is a mistake somewhere.

Hope that helps. :smile:
(edited 9 years ago)
Original post by lazy_fish

Just to clarify, your answer was n>log⁡4+log⁡1015log⁡2+log⁡3+1 \displaystyle n > \frac {\log 4 + \log 10^{15}} {\log 2 + \log 3} +1

If so, there is a mistake somewhere.

Hope that helps. :smile:


Actually can't see he mistake :frown:


Posted from TSR Mobile
Original post by Jimmy20002012
...


I mean mistake reaching that answer.

Mine was n>log⁡2+15log⁡3+1 \displaystyle n > \frac {\log 2 + 15} {\log 3} +1

:smile:
Original post by lazy_fish
I mean mistake reaching that answer.

Mine was n>log⁡2+15log⁡3+1 \displaystyle n > \frac {\log 2 + 15} {\log 3} +1

:smile:


Why is it that


Posted from TSR Mobile
Original post by Jimmy20002012
...


Maybe show the working? :smile:

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you