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C2 logs exam question problem

kinda stuck on this question , any help?



Find the value for Y such that Log2Y= -3
Reply 1
Remember that

logab = c

is equivalent to

b = ac
Reply 2
thank you !
:biggrin:
Reply 3
Original post by hicky@picky
kinda stuck on this question , any help?



Find the value for Y such that Log2Y= -3


log2y=-3
log2y=log-3
ylog2=log-3
y=log-3/log2

hope this helps
Reply 4
Original post by dongonaeatu
log2y=-3
log2y=log-3
ylog2=log-3
y=log-3/log2

hope this helps


You could confuse the OP by posting something like that, I can't tell if you're joking.

Is it log(2y) or log2(y) as in log to the base 2? Just to note that if there's no base given, that implies base 10, which will mean your answer will be (10^-3 )/2.
Reply 5
Original post by dongonaeatu
log2y=-3
log2y=log-3
ylog2=log-3
y=log-3/log2

hope this helps


Just ignore this, OP.
Reply 6
Original post by dongonaeatu
log2y=-3
log2y=log-3
ylog2=log-3
y=log-3/log2

hope this helps


Don't mind, but your concepts of logs are very weak. I remember that you have asked a lot of question before about logs, first you need to get good at this yourself then answer others. Otherwise, you will just confuse the OP.
Reply 7
Original post by raheem94
Don't mind, but your concepts of logs are very weak. I remember that you have asked a lot of question before about logs, first you need to get good at this yourself then answer others. Otherwise, you will just confuse the OP.


what did i do wrong in that
Reply 8
Original post by dongonaeatu
what did i do wrong in that


The question is: log2y=3 log_2y=-3

We will solve it by writing it as, 23=y 2^{-3} = y

I think i have explained this stuff to you before, right?

You can't do this:
log2y=-3
log2y=log-3
ylog2=log-3
y=log-3/log2


Remember if you have something like, logabc log_ab^c, then you can write it as, clogab clog_ab , but you are doing wrong.
Reply 9
Original post by dongonaeatu

log2y=-3
log2y=log-3



Original post by dongonaeatu
what did i do wrong in that



What made you think that you could simply log the RHS
Original post by TenOfThem
What made you think that you could simply log the RHS



Original post by raheem94
The question is: log2y=3 log_2y=-3

We will solve it by writing it as, 23=y 2^{-3} = y

I think i have explained this stuff to you before, right?

You can't do this:
log2y=-3
log2y=log-3
ylog2=log-3
y=log-3/log2


Remember if you have something like, logabc log_ab^c, then you can write it as, clogab clog_ab , but you are doing wrong.


i may have to look over logs again.
Reply 11
Original post by dongonaeatu
i may have to look over logs again.


You NEED to have a look, your concepts are still unclear.
Original post by raheem94
You NEED to have a look, your concepts are still unclear.


i think i'm going to fail c2
Reply 13
Original post by dongonaeatu
i think i'm going to fail c2


Don't be pessimistic, you still have time to improve.
Reply 14
dongonaeatu, you just gotta write out the 4 or 5 log rules out next to each other and rote memorise them one by one. You gotta have them all there on the same page so you don't get them mixed up.

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