Maths, Addmaths and Statistics: Post your doubts here!

[marshall mathers]

So does this mean I answered it correctly? It's the same as your final answer (7P1 x 6P2 = 7P3)!

Yes mate. Way to go dude!

 

[Hamza Iftikhar 242]

After the initiative taken by Math_angel in the IGCSE section, I have made this topic for students of O'levels. You can post any question or any problem here and me along with other XPF users will try to help you as soon as possible. This topic is a feasible option as it won't bring a hassle of making new topics for every question and it will also allow the helpers to not miss any of your questions and have one common topic for answering your Mathematics queries.

So start posting your problems.

hi guys, i am seriously stuck in this question that lg 2=m find log base8 5 in terms of m

 

[marshall mathers]

hi guys, i am seriously stuck in this question that lg 2=m find log base8 5 in terms of m

Could you rephrase the question because it seems quite ambiguous? Is this what you asked: given lg 2=m, find log 8 5 in terms of m??( log 5 to base 8)

 

[Hamza Iftikhar 242]

Could you rephrase the question because it seems quite ambiguous? Is this what you asked: given lg 2=m, find log 8 5 in terms of m??( log 5 to base 8)

i meant log of 8 to base 5

 

[marshall mathers]

Well the closest i can get to this answer is 3m/lg 5. What i did is expressing log 8 to base 5 to a common base such as 10. This gave me lg 8/lg 5. Then i wrote the lg 8 as 3lg2 and got the answer 3m/lg 5. But am definitely sure this is not the answer because you have to express the whole in terms of m. Sorry couldn't get any closer unless if it is m= log 2 to base 5

 

[Muniba Suhail]

hey, I've found the papers of may/june 2012 of add-math, but can't find their marking schemes, can anyone please suggest a site to download them from.

 

[Steel Arm]

hey, I've found the papers of may/june 2012 of add-math, but can't find their marking schemes, can anyone please suggest a site to download them from.

http://olevel.sourceforge.net/papers/4037/
check it here

 

[Steel Arm]

Hey can anyone please solve this for me...
2^(2x) multiply by 5^(x+1) is equal to 7

please guys help!!
p.s. must use logarithms!

 

[Muniba Suhail]

http://olevel.sourceforge.net/papers/4037/
check it here

thank you so much!

 

[Muniba Suhail]

Hey can anyone please solve this for me...
2^(2x) multiply by 5^(x+1) is equal to 7

please guys help!!
p.s. must use logarithms!

2^2x, is that 2 divided by 2x? multiplied with 5 divided by (x+1)?

 

[Steel Arm]

2^2x, is that 2 divided by 2x? multiplied with 5 divided by (x+1)?

no its 2 power 2x

 

[Muniba Suhail]

no its 2 power 2x

i have no idea to solve it now.

 

[Steel Arm]

i have no idea to solve it now.

no prob.....i think i know

 

[Muniba Suhail]

no prob.....i think i know

how? how are we going to solve it?

 

[Steel Arm]

how? how are we going to solve it?

okay - 5^(x+1) = 5^x X 5^1
therefore
you can simplify the equation to
2^(2x) X 5^x X 5^1 = 7
you send the 5^1 on the RHS and apply log
lg2^(2x) + lg5^x = lg(7/5)
you send the power x in front
xlg4+xlg5=lg(7/5)
x is common so you take it out and multiply the logs
x(lg20)=lg(7/5)
x=((lg(7/5))/lg20)
the ans is then 0.112

i know sounds complicated and difficult to read but..... thats it

 

[Miss_Z]

Somebody solve this for me, please.

5 (sin^2) y + 9 cos y - 3 = 0. ( 0 < y < 360 degrees)

Include full details of each step if possible, tyvm. God Bless

 

[Steel Arm]

Somebody solve this for me, please.

5 (sin^2) y + 9 cos y - 3 = 0. ( 0 < y < 360 degrees)

Include full details of each step if possible, tyvm. God Bless

firstly i would change sin^2y with 1-cos^2y
then by expanding you will come to
5cos^2y - 9cosy - 2 = 0
then you let cosy = x
then you solve the equation to get
either x= 2 or x=-1.5
then you reject 2 and work with -1.5
cosy=-1.5
you solve that easily then

 

[Steel Arm]

firstly i would change sin^2y with 1-cos^2y
then by expanding you will come to
5cos^2y - 9cosy - 2 = 0
then you let cosy = x
then you solve the equation to get
either x= 2 or x=-1.5
then you reject 2 and work with -1.5
cosy=-1.5
you solve that easily then

SRRY that shoud be 1/5 and not 1.5

 

[Miss_Z]

SRRY that shoud be 1/5 and not 1.5

Ah right, thanks a lot. I did the same steps but didn't get the right answers, so I didn't know where my mistake was. I had actually taken o.2 instead of -0.2. Thanks once more.


Cheers.

P.s: Your signature states one of the truest things in the universe LOL.

 

[Steel Arm]

Ah right, thanks a lot. I did the same steps but didn't get the right answers, so I didn't know where my mistake was. I had actually taken o.2 instead of -0.2. Thanks once more.


Cheers.

P.s: Your signature states one of the truest things in the universe LOL.

No probs. And THX