The Student Room Group

Scroll to see replies

Reply 20
Original post by dongonaeatu
dont understand the second step


I have used the change of base law in order to make both of the bases the same so I can then combine the logs into one single log.

The change of base law is as follows:

logax = logbx/logba

^ Learn that formula.

In this case we had:

log2x + log4x = 2

a = 4
b = 2

So it turned into:

log2x + log2x/log24 = 2
Reply 21
Original post by Joshmeid
log2x + log4x = 2

log2x + log2x/log24 = 2

log2x + log2x/2 = 2

2log2x + log2x = 4

3log2x = 4

log2x^3 = 4

x^3 = 16

x = 2.52(3sf)

Ask if you are confused by anything.


Please don't post full solutions, they are considered a last resort.
Reply 22
Original post by Joshmeid
I have used the change of base law in order to make both of the bases the same so I can then combine the logs into one single log.

The change of base law is as follows:

logax = logbx/logba

^ Learn that formula.

In this case we had:

log2x + log4x = 2

a = 4
b = 2

So it turned into:

log2x + log2x/log24 = 2


thanks i go it :smile:
Reply 23
Original post by avtar95
could u put that in letters


According to the change of base rule logab \displaystyle log_ab can be written as logcblogca \displaystyle \frac{log_cb}{log_ca}

In the question it was log4x \displaystyle log_4x
So here a=4 and b=x \displaystyle a=4 \text{ and } b=x

Lets use the change of base rule, we want the base of logs to be 2 hence let c=2 \displaystyle c=2
log4x=log2xlog24 \displaystyle log_4x = \frac{log_2x}{log_24}

Do you get it?
Original post by raheem94
According to the change of base rule logab \displaystyle log_ab can be written as logcblogca \displaystyle \frac{log_cb}{log_ca}

In the question it was log4x \displaystyle log_4x
So here a=4 and b=x \displaystyle a=4 \text{ and } b=x

Lets use the change of base rule, we want the base of logs to be 2 hence let c=2 \displaystyle c=2
log4x=log2xlog24 \displaystyle log_4x = \frac{log_2x}{log_24}

Do you get it?


kinda its quite complicated though
Reply 25
Original post by dongonaeatu
is that rule just for log2x and log4x i've never been taught that


Remember logab+logac=loga(bc) \displaystyle log_ab + log_ac = log_a(bc)

So log2x+log4x \displaystyle log_2x + log_4x cannot be written in that way as both have different bases.
Reply 26
Original post by raheem94
Please don't post full solutions, they are considered a last resort.


thanks i go it :smile:
Reply 27
Original post by dongonaeatu
kinda its quite complicated though


It isn't complicated with a bit of practice you will be fine with it.
Original post by raheem94
Remember logab+logac=loga(bc) \displaystyle log_ab + log_ac = log_a(bc)

So log2x+log4x \displaystyle log_2x + log_4x cannot be written in that way as both have different bases.


are the bases the a's so the number in front of the log
Reply 29
Original post by dongonaeatu
are the bases the a's so the number in front of the log


The a's are the bases.
Original post by raheem94
It isn't complicated with a bit of practice you will be fine with it.


so loga 5 +loga 4= loga20 as the base is the same (a)

but loga5+ logb4= not log20 as different bases what would i do in this instance
Reply 31
Original post by dongonaeatu
so loga 5 +loga 4= loga20 as the base is the same (a)

but loga5+ logb4= not log20 as different bases what would i do in this instance


Yes, loga5+loga4=loga(5×4)=loga(20) \displaystyle log_a5 + log_a4 = log_a(5 \times 4) = log_a(20)


And loga5+logb4log20 \displaystyle log_a5+log_b4 \not= log20

For solving such questions you need to put both on the same base. Lets change the base of logb4 \displaystyle log_b4 to a \displaystyle a

logb4=loga4logab \displaystyle log_b4 = \frac{log_a4}{log_ab}

So now the equation becomes, loga5+loga4logab \displaystyle log_a5 + \frac{log_a4}{log_ab}
You can see that now all the logarithms have the same base.

Do i make any sense?
Original post by raheem94
Yes, loga5+loga4=loga(5×4)=loga(20) \displaystyle log_a5 + log_a4 = log_a(5 \times 4) = log_a(20)


And loga5+logb4log20 \displaystyle log_a5+log_b4 \not= log20

For solving such questions you need to put both on the same base. Lets change the base of logb4 \displaystyle log_b4 to a \displaystyle a

logb4=loga4logab \displaystyle log_b4 = \frac{log_a4}{log_ab}

So now the equation becomes, loga5+loga4logab \displaystyle log_a5 + \frac{log_a4}{log_ab}
You can see that now all the logarithms have the same base.

Do i make any sense?


did u use the change of base loga^b=logc to the b/logc to the a

and u make C the base ur changing it to
Reply 33
Original post by dongonaeatu
did u use the change of base loga^b=logc to the b/logc to the a

and u make C the base ur changing it to


I am not understanding your question.

I used the change of base rule to convert logb4 \displaystyle log_b4 to loga4logab \displaystyle \frac{log_a4}{log_ab}
Original post by raheem94
I am not understanding your question.

I used the change of base rule to convert logb4 \displaystyle log_b4 to loga4logab \displaystyle \frac{log_a4}{log_ab}


do i have to learn the change of base formula

and the C in the equation is the log of what u are converting it to so if i want it in base of a i have the C as a
Reply 35
Original post by dongonaeatu
do i have to learn the change of base formula

and the C in the equation is the log of what u are converting it to so if i want it in base of a i have the C as a


Yes, you have to learn the change of base formula.

And yes the C in my equation is 'a', and your concept about this is correct. I wanted the log to have the base 'a' hence i set 'C' to 'a'.
Original post by raheem94
Yes, you have to learn the change of base formula.

And yes the C in my equation is 'a', and your concept about this is correct. I wanted the log to have the base 'a' hence i set 'C' to 'a'.


do i get given the formula in the exam and how many hours of maths do u do a day u are amazing
Reply 37
Original post by dongonaeatu
do i get given the formula in the exam and how many hours of maths do u do a day u are amazing


I don't know if the formula will be given or not, but this is a simple formula. Just understand it, don't try to mug it up.

I have already finished A-Level maths(C1-4, M1-2), with an A* :smile:, now i am doing further maths, so C2 is simple for me. I don't work too much, A-Level maths isn't hard. Just go through the chapters in the book and you will be fine with it. Try to understand things, do you use a book?
Original post by raheem94
I don't know if the formula will be given or not, but this is a simple formula. Just understand it, don't try to mug it up.

I have already finished A-Level maths(C1-4, M1-2), with an A* :smile:, now i am doing further maths, so C2 is simple for me. I don't work too much, A-Level maths isn't hard. Just go through the chapters in the book and you will be fine with it. Try to understand things, do you use a book?


wow A* thats great well done. i have a green edxcel textbook and a c2 resource pack and access to past papers and marked solutions
Reply 39
Original post by dongonaeatu
wow A* thats great well done. i have a green edxcel textbook and a c2 resource pack and access to past papers and marked solutions


I also used the green edexcel text books, they are great. I self-studied all units using them, they explain really well. Have you read the chapters properly?