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c2 logs
Hi i dont get this question:
given that p=logq(pq), express in terms of p:
a) logq2
b) logq(8q)
The answers are:
a) 0.25p
b) 0.74p + 1
But I don't get how to do it?
Thanks
given that p=logq(pq), express in terms of p:
a) logq2
b) logq(8q)
The answers are:
a) 0.25p
b) 0.74p + 1
But I don't get how to do it?
Thanks
Reply 4
Where is this question from? I got to the same point before getting stuck 
If this is like part c) of some question then there may be information at the start which could help
If this is like part c) of some question then there may be information at the start which could helpReply 5
I think you've got the question wrong. I did a similar question the other day, it was like this;
Given that p=logq16 express in terms of p,
a logq2
b logq(8q)
a p=logq16
p=4logq2
logq2=p/4
b logq(8q)
p=logq16
p=logq8 + logqq
3logq2 + 1
3(p/4) + 1
therefore logq(8q) = 3p/4 + 1
I hope its not too confusing to understand, if it is look at my attachment.
Given that p=logq16 express in terms of p,
a logq2
b logq(8q)
a p=logq16
p=4logq2
logq2=p/4
b logq(8q)
p=logq16
p=logq8 + logqq
3logq2 + 1
3(p/4) + 1
therefore logq(8q) = 3p/4 + 1
I hope its not too confusing to understand, if it is look at my attachment.
Reply 6
Phew, I'm not the only one who got stuck! I was up 'til 2 am trying everything.
Reply 7
maffsman
OP
thanks I get it now!
Did you get that out of the Hienenman Core 2 book because that's where I got it from. I seem to be stuck on most of the Mixed Exercise section for logs...could any one help me on any of them?
It's because the chapter doesn't have any questions like the ones that are in the Mixed Exercise, and that's all I have to revise from and my exam is tomorrow!!!!
One question is:
4a) Given that log3x=2 determine the value of x
b) Calculate the value of y for which 2log3y - log3(y+4)=2
c) Calculate the values of z for which log3z=4logz3
I just don't understand what to do. The answers are:
a) 9
b) 12
c) 1/9, 9 <------ for this question I got 9 but I didn't get any other values
Did you get that out of the Hienenman Core 2 book because that's where I got it from. I seem to be stuck on most of the Mixed Exercise section for logs...could any one help me on any of them?
It's because the chapter doesn't have any questions like the ones that are in the Mixed Exercise, and that's all I have to revise from and my exam is tomorrow!!!!
One question is:
4a) Given that log3x=2 determine the value of x
b) Calculate the value of y for which 2log3y - log3(y+4)=2
c) Calculate the values of z for which log3z=4logz3
I just don't understand what to do. The answers are:
a) 9
b) 12
c) 1/9, 9 <------ for this question I got 9 but I didn't get any other values
maffsman
thanks I get it now!
Did you get that out of the Hienenman Core 2 book because that's where I got it from. I seem to be stuck on most of the Mixed Exercise section for logs...could any one help me on any of them?
It's because the chapter doesn't have any questions like the ones that are in the Mixed Exercise, and that's all I have to revise from and my exam is tomorrow!!!!
One question is:
4a) Given that log3x=2 determine the value of x
b) Calculate the value of y for which 2log3y - log3(y+4)=2
c) Calculate the values of z for which log3z=4logz3
I just don't understand what to do. The answers are:
a) 9
b) 12
c) 1/9, 9 <------ for this question I got 9 but I didn't get any other values
Did you get that out of the Hienenman Core 2 book because that's where I got it from. I seem to be stuck on most of the Mixed Exercise section for logs...could any one help me on any of them?
It's because the chapter doesn't have any questions like the ones that are in the Mixed Exercise, and that's all I have to revise from and my exam is tomorrow!!!!
One question is:
4a) Given that log3x=2 determine the value of x
b) Calculate the value of y for which 2log3y - log3(y+4)=2
c) Calculate the values of z for which log3z=4logz3
I just don't understand what to do. The answers are:
a) 9
b) 12
c) 1/9, 9 <------ for this question I got 9 but I didn't get any other values
for part a) write it in indices form to find x.
for part b) use one of the log rules to combile the left hand rule and do the same thing as in part a). hope that helps
Reply 9
maffsman
OP
I've done a) now thanks but with b) I got to here:
log3y2/(y+4)=2
y2/y+4=23
y2=8y+32
y2-8y-32=0
(y )(y )=0 It doesn't go into brackets does it?
log3y2/(y+4)=2
y2/y+4=23
y2=8y+32
y2-8y-32=0
(y )(y )=0 It doesn't go into brackets does it?
Reply 10
maffsman
4a) You've probably done this one. x = 9
maffsman
b) Calculate the value of y for which 2log3y - log3(y+4)=2
Third law of logs:
log3(y2) - log3(y + 4) = 2
Second law of logs (division):
log3(y2 / (y + 4)) = 2
32 = 9 so:
y2 / (y + 4) = 9
Then multiply:
y2 = 9y + 36
Rearrange for a quadratic polynomial:
y2 - 9y - 36 = 0
Then solve as a quadratic. Answers: 12 and -3, but only 12 works because of log restrictions (I think).
c) Calculate the values of z for which log3z=4logz3
Third law of logs:
log3z=logz34
34 = 81. Then express the above as an exponential:
zlog3z = 81
Then log3 both sides of the equation:
log3zlog3z = log381
Third law of logs, and solve RHS:
log3z * log3z = 4
Express LHS as a square
(log3z)2 = 4
Swap the square on the LHS for a square root on the RHS
log3z = 40.5
Simplify RHS (remember that the square root of something will be + or - the answer)
log3z = + or - 2
Then solve for both + or - 2. Answers are 1/9 and 9
Reply 11
maffsman
OP
Thanks that really helps, but there was something for part b that you did which was unfamiliar to me:
So when you take the logs away and just work with the numbers, do you take the base of the log (ie 3) and put that to the power of the RHS (ie 2)
So you get 32?
Because I've only come across this once which was in an exam question, but the base was 2 and the RHS=2 so when they did 22 I assumed that they did the RHS to the power of the base, but then that isn't what you did.
Don't know if you understand what I mean by all that!! But if you do could you please just confirm the rule. Thanks
PlaystationStudies
log3(y2 / (y + 4)) = 2
32 = 9 so:
y2 / (y + 4) = 9
log3(y2 / (y + 4)) = 2
32 = 9 so:
y2 / (y + 4) = 9
So when you take the logs away and just work with the numbers, do you take the base of the log (ie 3) and put that to the power of the RHS (ie 2)
So you get 32?
Because I've only come across this once which was in an exam question, but the base was 2 and the RHS=2 so when they did 22 I assumed that they did the RHS to the power of the base, but then that isn't what you did.
Don't know if you understand what I mean by all that!! But if you do could you please just confirm the rule. Thanks
Reply 12
Exactly 
If in doubt, remember that
log10100 = 2
What power of 10 equals 100? The answer is 2.
And...
log3(y2 / (y + 4)) = 2
...is the same as...
32 = (y2 / (y + 4))
If in doubt, remember that
log10100 = 2
What power of 10 equals 100? The answer is 2.
And...
log3(y2 / (y + 4)) = 2
...is the same as...
32 = (y2 / (y + 4))
Reply 13
maffsman
OP
Ahaa lol I get it now!! Thanks.
Reply 14
maffsman
OP
Right another question!!:
Solve, giving your answers as exact fractions, the simultanious equations:
8y=42x+3
log2y=log2x+4
Solve, giving your answers as exact fractions, the simultanious equations:
8y=42x+3
log2y=log2x+4
Reply 15
The same question is being asked in this thread: http://thestudentroom.co.uk/showthread.php?t=393163
Log the first equation, and try and isolate either x or y so that you can feed it into the second equation.
Log the first equation, and try and isolate either x or y so that you can feed it into the second equation.
Reply 16
maffsman
OP
I still can't do it do you have the working cos I did what you said but keep seem to be getting the wrong answer.
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