Algebraic manipulation (q8 2022 paper 1)

OP

Hi everyone thank you so much for your responses! I know a lot of you are confused as to what I have done after the second line. Here's what I did:
- 3rd line: divided everything by 0.4 ln (t+1)
- 4th line: factored out 0.5 in the fraction to get (25-t) in the numerator
- 5th line : cancelled out the 0.4

Many people have said to equal the equation to 0- I have already done that I just didn't make that obvious on the whiteboard (my bad)

This is for part b btw and I am really struggling to rearrange the equation for the required form

Reply 8

OP

Original post by mathsstudent222

I posted solution but daddy moderator deleted it

ahhhhh oh well. I have seen the solution to it on youtube but I was wondering if I would get anywhere with the way I have done it ? I don't understand where I have went wrong

Reply 9

old_engineer

11

Original post by leoishush

ahhhhh oh well. I have seen the solution to it on youtube but I was wondering if I would get anywhere with the way I have done it ? I don't understand where I have went wrong

You're not necessarily wrong, but you have headed off in a slightly strange direction. What I would suggest is:

* Go back to line 2
* Start the line with "0 =" rather than just "="
* Concentrate on making "t" the subject of the equation

Reply 10

Original post by leoishush

ahhhhh oh well. I have seen the solution to it on youtube but I was wondering if I would get anywhere with the way I have done it ? I don't understand where I have went wrong

If I was doing it I would have worked back from the answer 1 step and tried to prove
(t+1) (1 + ln(t+1)) = 26

You dont seem to set dv/dt to zero. If you did youd get
0.4 ln(t+1) = (10 - 0.4t) / (t+1)
It should be straightforward to get the above answer by multiplying through and rearranging?

Reply 11

OP

Original post by mqb2766

If I was doing it I would have worked back from the answer 1 step and tried to prove
(t+1) (1 + ln(t+1)) = 26

You dont seem to set dv/dt to zero. If you did youd get
0.4 ln(t+1) = (10 - 0.4t) / (t+1)
It should be straightforward to get the above answer by multiplying through and rearrangin

ahhhh right. I have now tried doing that but for some reason i am still not getting the right answer
(line 3 I cancelled out the 0.4, line 4 I swapped t+1and ln(t+1)

Reply 12

I would have multiplied through, but add zero = 1 - 1 to the numerator should get you about there?

Reply 13

OP

Original post by mqb2766

I would have multiplied through, but add zero = 1 - 1 to the numerator should get you about there?

sorry i am confused

Reply 14

Original post by leoishush

sorry i am confused

Multiply through and add 1 - 1 to the right gives
(t+1) ln(t+1) = 26 - t - 1
....

Reply 15

OP

Original post by mqb2766

Multiply through and add 1 - 1 to the right gives
(t+1) ln(t+1) = 26 - t - 1
....

for some reason I am still stuck with 25 in my working

Reply 16

Original post by leoishush

for some reason I am still stuck with 25 in my working

Multiply through you get
(t+1)ln(t+1) = 25 - t
you want 26, you also want a factor t+1 on the right. So add 0 = 1 - 1 to the right to get
(t+1)ln(t+1) = 25 + 1 - t - 1 = 26 - (t+1)
take the (t+1) on the right over, factorise and divide ...

For me there were a few ways to proceed but to rearrange the desired relationship as in the previous post, then notice that
10/0.4 = 25
and we needed 26 and an extra t+1 factor pretty much got it done.

(edited 10 months ago)

Reply 17

OP

Original post by mqb2766

Multiply through you get
(t+1)ln(t+1) = 25 - t
you want 26, you also want a factor t+1 on the right. So add 0 = 1 - 1 to the right to get
(t+1)ln(t+1) = 25 + 1 - t - 1 = 26 - (t+1)
take the (t+1) on the right over, factorise and divide ...

For me there were a few ways to proceed but to rearrange the desired relationship as in the previous post, then notice that
10/0.4 = 25
and we needed 26 and an extra t+1 factor pretty much got it done.

ohhhhhhhh thank you so much!

Reply 18

Original post by leoishush

ohhhhhhhh thank you so much!

For questions like this, if necessary try and work back a step or two. As there was t on both sides of the desired relationship, for me it indicated that multiplying through back from the final solution and just think what bits you have and how to rearrange / factorise them was important before really putting pen to paper.

(edited 10 months ago)

Reply 19

OP