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logs - c2

hi can someone tell me please the formula if you need to make the same log bases such as logbase3(2-3x)=logbase9(6x2-19x+2)

how would i make the same bases 3 what is formula thanks :wink:

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Reply 1
Original post by otrivine
hi can someone tell me please the formula if you need to make the same log bases such as logbase3(2-3x)=logbase9(6x2-19x+2)

how would i make the same bases 3 what is formula thanks :wink:


logab=logcblogca \displaystyle log_ab = \frac{log_cb}{log_ca}

Example 1:

log39 \displaystyle log_39
If we want to make it to the base 5.
log39=log59log53 \displaystyle log_39 = \frac{log_59}{log_53}

Example 2:

log37 \displaystyle log_37
We want to convert it to the base 7.
log37=log77log73=1log73 \displaystyle log_37 = \frac{log_77}{log_73} =\frac1{log_73}
Reply 2
Original post by raheem94
logab=logcblogca \displaystyle log_ab = \frac{log_cb}{log_ca}

Example 1:

log39 \displaystyle log_39
If we want to make it to the base 5.
log39=log59log53 \displaystyle log_39 = \frac{log_59}{log_53}

Example 2:

log37 \displaystyle log_37
We want to convert it to the base 7.
log37=log77log73=1log73 \displaystyle log_37 = \frac{log_77}{log_73} =\frac1{log_73}

one question p=logbaseq16 how can you make the base q subject?
Reply 3
loga(b)=clog_a(b) = c


b=acb = a^c


a=b1c=bca = b^{\frac{1}{c}} = \sqrt[c]{b}
Reply 4
Original post by TenOfThem
loga(b)=clog_a(b) = c


b=acb = a^c


a=b1c=bca = b^{\frac{1}{c}} = \sqrt[c]{b}


do we need to know that for c2 cause i am solving this one where it says
p=logbaseq16 find in terms of P
a)logbaseq2
Reply 5
Original post by otrivine
do we need to know that for c2 cause i am solving this one where it says
p=logbaseq16 find in terms of P
a)logbaseq2


p=logq16 p=log_q16

Remember, 16=24 16 = 2^4

Example:

p=logm4 p=log_m4
You are asked to find logm2 log_m2 in terms of pp.

So here is how we will do it,
logm2=logm4=logm412=12logm4 log_m2 = log_m\sqrt4 = log_m4^{\frac12}= \frac12 log_m4.
We know logm4log_m4 equals p p so we can write 12logm4 \frac12 log_m4 as 12p \frac12 p
Reply 6
Original post by otrivine
one question p=logbaseq16 how can you make the base q subject?


p=logq16 p=log_q 16

To make it the subject, use the rule, loga(b)=c log_a(b) = c can be written as ac=b a^c =b

Compare, p=logq16 p=log_q 16 with loga(b)=c log_a(b) = c .
Here, p=c  q=a  16=b p= c \ \ q=a \ \ 16 = b

So now write it in the way ac=b a^c=b , sub in the above data, we will get, qp=16 q^p = 16

Now to remove the power of p p from q q , follow the following step:
qp=16qp×1p=161pq=161p[br][br] q^p = 16 \\ q^{p \times \frac1p} = 16^{\frac1p} \\ q = 16^{\frac1p}[br][br]
(edited 11 years ago)
Reply 7
Original post by otrivine
do we need to know that for c2


It is the core definition of logs


Original post by otrivine
cause i am solving this one where it says
p=logbaseq16 find in terms of P
a)logbaseq2



A Raheem said this does not require anything beyond knowing that 2^4 = 16
Reply 8
Original post by TenOfThem
It is the core definition of logs





A Raheem said this does not require anything beyond knowing that 2^4 = 16


thanks and last question ten of them this question
http://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/June%202011%20-%20QP/6664_01_que_20110526.pdf

8)a) to do that how ? i got 2xpower3 for volume then what i do?
Reply 9
Original post by otrivine
thanks and last question ten of them this question
http://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/June%202011%20-%20QP/6664_01_que_20110526.pdf

8)a) to do that how ? i got 2xpower3 for volume then what i do?


Let 'm' be the other side of the cuboid.

Remember, Volume =2x×x×m \text{Volume} \ = 2x \times x \times m

We know volume equals 81, find an equation for 'm'.

L = 4(2x) + 4x +4m
Reply 10
Original post by raheem94
Let 'm' be the other side of the cuboid.

Remember, Volume =2x×x×m \text{Volume} \ = 2x \times x \times m

We know volume equals 81, find an equation for 'm'.

L = 4(2x) + 4x +4m

yep got the answer thanks and why did u put m and not x
ANd for C2 what volume and lengths and area do we need to know to be able to apply like these questions?
Reply 11
Original post by otrivine
yep got the answer thanks and why did u put m and not x
ANd for C2 what volume and lengths and area do we need to know to be able to apply like these questions?


How do you know the other length will be 'x'?

Another length is specified as 'x' so i should now choose a different variable.

Just learn the formulas given in the book, to be honest, this part a of the question was o-level stuff.
Reply 12
Original post by raheem94
How do you know the other length will be 'x'?

Another length is specified as 'x' so i should now choose a different variable.

Just learn the formulas given in the book, to be honest, this part a of the question was o-level stuff.

thanks you are the best
but i did not get the question on logs where we said let p=logbaseq16 find in terms of p logbaseq2?
Reply 13
Original post by otrivine
thanks you are the best
but i did not get the question on logs where we said let p=logbaseq16 find in terms of p logbaseq2?


I don't have any idea of explaining you without doing a solution, so i will have to go for it.

p=logq16=logq24=4logq2    p4=logq2 p=log_q 16 = log_q2^4 = 4log_q2 \implies \dfrac{p}4 = log_q2

So logq2=p4 \boxed{ log_q2 = \dfrac{p}{4} }

You look to be posting a lot, so i think you should consider learning LaTex.
(edited 11 years ago)
Reply 14
Original post by raheem94
I don't have any idea of explaining you without doing a solution, so i will have to go for it.

p=logq16=logqi24=4logq2    p4=logq2 p=log_q 16 = log_qi2^4 = 4log_q2 \implies \dfrac{p}4 = log_q2

So logq2=p4 \boxed{ log_q2 = \dfrac{p}{4} }

You look to be posting a lot, so i think you should consider learning LaTex.


thanks ok http://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/June%202009/6664_01_que_20090605.pdf
question 9)a) i did h=300/2r
and sub into the the area but is not working?
Reply 15
Original post by otrivine
thanks ok http://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/June%202009/6664_01_que_20090605.pdf
question 9)a) i did h=300/2r
and sub into the the area but is not working?


You need to work hard on your basics mate.

A=12r2θ=12r2 A = \frac12 r^2 \theta = \frac12 r^2
V=A×h V = A \times h
We know V= 300, find 'h' in terms of r^2.

Then find the surface area, i am assuming you are clear with the meaning of surface area.
Reply 16
Original post by raheem94
You need to work hard on your basics mate.

A=12r2θ=12r2 A = \frac12 r^2 \theta = \frac12 r^2
V=A×h V = A \times h
We know V= 300, find 'h' in terms of r^2.

Then find the surface area, i am assuming you are clear with the meaning of surface area.


how can i improve my basics for C2 please tell me raheem

what does A and H stand for in volume
(edited 11 years ago)
Reply 17
Original post by otrivine
how can i improve my basics for C2 please tell me raheem

what does A and H stand for in volume


There is no magic wand to improve yourself.

I just used to read all the examples and solve all the questions from the book.

A stands for area.

The 'H' is 'h', and it is labelled in the diagram.
Reply 18
Original post by raheem94
There is no magic wand to improve yourself.

I just used to read all the examples and solve all the questions from the book.

A stands for area.

The 'H' is 'h', and it is labelled in the diagram.

shall i do all the mixed exercise?
Reply 19
Original post by otrivine
shall i do all the mixed exercise?


:yep:

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