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Re: Nairaland Mathematics Clinic by Odehfamily(m): 5:49pm On Jan 03, 2013
Youngsage:
x + y =5 -------- eqn i.
x^x + y^y=31 ----- eqn ii.
From equatns i & ii, take the log of both sides
(xlog^x + ylog^y) = log 31------ eqn iii.
logx + logy = log 5 ------------ eqn iv.
xylogxy= log 31 -------- eqn v.
log xy = log 5 --------- eqn vi.
(Using elimination method);
xylog5 = log31
xy = log 31/log 5 =6.2 approx. 6.

Recall, from eqn i, x + y = 5.
So x = 6/y or y = 6/x.
Therefore x=2 when y=3.
ans: x=2, and y=3.

Excellent
Re: Nairaland Mathematics Clinic by ositadima1(m): 5:57pm On Jan 03, 2013
Odehfamily:

Excellent

Is log(x+y)=log5 same as logx+logy=log5

1 Like

Re: Nairaland Mathematics Clinic by Nobody: 5:57pm On Jan 03, 2013
Nice one.bookmarked n subscribing!
Re: Nairaland Mathematics Clinic by Youngsage: 5:58pm On Jan 03, 2013
Wow, been off for a while and 7 pages already! Never knew dere were so many 'gurus' on NL.
@O.P gud job you doing here, i neva had much interest in maths b4 nt until recently, i must say i admire ur mathematical dexterity, wat can one do to master topics in maths? hope to be a guru lyk u someday..., @all lets kip the comments flowing so dis thread can stay longer on frontpage.
Let me go over wat i've missd so far previously...be back soon!
Re: Nairaland Mathematics Clinic by Nobody: 6:01pm On Jan 03, 2013
ositadima1:

Is log(x+y)=log5 same as logx+logy=log5
no, log(xy)=log x + log y
Re: Nairaland Mathematics Clinic by ladokuntlad(m): 6:09pm On Jan 03, 2013
coolexy2: (1)what is the real application of mathematics in our live?

(2) prove that:
a. Sin^3x = 1/4(3sinx - sin3x
b. Find 2nd derivative of:
y = cos^3x
Re: Nairaland Mathematics Clinic by Youngsage: 6:09pm On Jan 03, 2013
CHYKMAN:

Log(x+y) = Log 5.. Is it the same as log x + log y = Log 5?? Cross check your solvings..
Previously, I said taking the log of both sides, i.e log of L.H.S and R.H.S of 'equals to' sign. From x + y = 5, this would imply that: log(x+y) = log5, not logx + logy = log5 as this would have a different implication. Thanks for your observation Chykman!
Re: Nairaland Mathematics Clinic by pekelepekele(m): 6:09pm On Jan 03, 2013
Gugurus (popcorns)cheesy
Re: Nairaland Mathematics Clinic by ositadima1(m): 6:10pm On Jan 03, 2013
Beretta92: no, log(xy)=log x + log y

He still got a correct answer,
Re: Nairaland Mathematics Clinic by ladokuntlad(m): 6:19pm On Jan 03, 2013
[quote author=ladokuntlad][/quote]

Re: Nairaland Mathematics Clinic by Odehfamily(m): 6:26pm On Jan 03, 2013
ositadima1:

Is log(x+y)=log5 same as logx+logy=log5
Open the bracket,by multiplying log by x and y. U wil hav logx+logy=log5
Re: Nairaland Mathematics Clinic by ositadima1(m): 6:31pm On Jan 03, 2013
Odehfamily:
Open the bracket,by multiplying log by x and y. U wil hav logx+logy=log5

U cant multiply log, its neither a number nor a variable.

Instead: log(xy) = logx + logy
Re: Nairaland Mathematics Clinic by ladokuntlad(m): 6:32pm On Jan 03, 2013
coolexy2: (1)what is the real application of mathematics in our live?

(2) prove that:
a. Sin^3x = 1/4(3sinx - sin3x
b. Find 2nd derivative of:
y = cos^3x

Re: Nairaland Mathematics Clinic by ositadima1(m): 6:36pm On Jan 03, 2013
^^^ dat is d first derivative
Re: Nairaland Mathematics Clinic by Leqsy(m): 6:37pm On Jan 03, 2013
What is a reducible element and are 3,4and 7 reducible in z(i).z indicating set of integers and i rep a complex no i.e sqrt of-1
Re: Nairaland Mathematics Clinic by Odehfamily(m): 6:51pm On Jan 03, 2013
Youngsage:
Previously, I said taking the log of both sides, i.e log of L.H.S and R.H.S of 'equals to' sign. From x + y = 5, this would imply that: log(x+y) = log5, not logx + logy = log5 as this would have a different implication. Thanks for your observation Chykman!
plz youngsaga! Help me explain the last step. Xy=6 then x=6/y and y=6/x. Plz explain that "RECALL THAT X+Y=5"
Re: Nairaland Mathematics Clinic by Kessyy2k4(m): 7:08pm On Jan 03, 2013
Youngsage:
x + y =5 -------- eqn i.
x^x + y^y=31 ----- eqn ii.
From equatns i & ii, take the log of both sides
(xlog^x + ylog^y) = log 31------ eqn iii.
log(x+y) = log 5 ------------ eqn iv.
xylogxy= log 31 -------- eqn v.
log xy = log 5 --------- eqn vi.
(Using elimination method);
xylog5 = log31
xy = log 31/log 5 =6.2 approx. 6.

Recall, from eqn i, x + y = 5.
So x = 6/y or y = 6/x.
Therefore x=2 when y=3.
ans: x=2, and y=3.
The falacy here is wen u say: xlogx+ylogy=xylogxy. Pls note xlogx+ylogy IS NOT=xylogxy. For example 2log3+4log5NOT=8log15. This has been verified. Buh i can absolutely agree with riches version of d solution by: 'try & error', this is d only accepted method. I.e x+y=5....(1)
x^x+y^y=31....(2)
from eqn (1), x=5-y & ... See riches solution. Thks
Re: Nairaland Mathematics Clinic by Barywhyte(m): 7:30pm On Jan 03, 2013
Odehfamily:
Open the bracket,by multiplying log by x and y. U wil hav logx+logy=log5

I have gone through this thread once and I saw that good pple have corrected this misconception. What u did above is a pure violation of fundamental principle of mathematics.

Log (x + y) can never be equal to logx + logy.

Logarithm is a function with a single definition and x + y is a single entity ok.

Tell me: can u say that sin (x + y) = sinx + siny? Of course not.

Agreed he got d correct answers but dubiously and in mathematics, your marks stop immediately a flaw is discovered and the examiner would not bother looking @ the rest of the solution

You guys r breaking d rules of mathematics.

Log (xy) = logx + logy : this is the first law of algebraic logarithm and is true for all positive natural numbers
Re: Nairaland Mathematics Clinic by awesomebenji(m): 7:31pm On Jan 03, 2013
@OP especially, simultaneous eqn in 3unknowns.
Given that:
a1x + b1y + c1z + d1 = 0
a2x + b2y + c2z + d2 = 0
a3x + b3y + c3z + d3 = 0
find x, y & z and leave d answer in matrix form.
Please, please & please, help me 2 solve dis as d lecturer has promised 2 bring it in d exam. Thanks.
My email add: olubanjisam@gmail.com
Re: Nairaland Mathematics Clinic by femi4: 7:34pm On Jan 03, 2013
ositadima1:

He still got a correct answer,
Sometimes in Mathematics, you can be fortunate to arrive at the right answers using wrong method depending on the variables involved
Re: Nairaland Mathematics Clinic by Time2talk(m): 7:40pm On Jan 03, 2013
[quote author=Richiez]This thread is the meeting point for nairaland maths gurus...i dare you all to ask me a question in mathematics without me having an answer to them, LETS START



Try solve this:
(X^2+Y^2)dx + 2XYdy = 0
Re: Nairaland Mathematics Clinic by YoungMT: 7:42pm On Jan 03, 2013
Help me on this que..the number of matches that would be played if 20 football clubs are to play in a league based on home and away,workings needed.
Re: Nairaland Mathematics Clinic by ositadima1(m): 7:51pm On Jan 03, 2013
Time2talk: Try solve this:
(X^2+Y^2)dx + (2XY)dy = 0
ODE, hmmm?

y=c-[x(X^2+Y^2)]/2XY
Where c is d arbitrary constant.
Do u want to c d steps?
Re: Nairaland Mathematics Clinic by Richiez(m): 7:57pm On Jan 03, 2013
Proflakes: x+cosx=2.154, find x

Proflakes: x+cosx=2.154, find x

SOLUTION
x+cosx=2.154.....(1)
we have to apply maclaurin's series for cosx= 1- x^2/2! + x^4/4!......
hence, x+ 1- x^2/2! + x^4/4!......=2.154
x^4/24- x^2/2 + x =2.154-1
x^4/24- x^2/2 + x =1.154
by factorization; x[x^3/24- x/2 + 1] = 1.154
therefore; x= 1.154 or x^3/24- x/2 + 1 = 1.154

with this we have succeeded in getting on value of x=1.154, solve for other values

1 Like

Re: Nairaland Mathematics Clinic by pawn89(m): 7:58pm On Jan 03, 2013
am a lover of maths, and i will willingly give mu input to any question asked regardless of the level
Re: Nairaland Mathematics Clinic by Horlalaykon: 8:06pm On Jan 03, 2013
.
Re: Nairaland Mathematics Clinic by ositadima1(m): 8:06pm On Jan 03, 2013
Richiez:



SOLUTION
x+cosx=2.154.....(1)
we have to apply maclaurin's series for cosx= 1- x^2/2! + x^4/4!......
hence, x+ 1- x^2/2! + x^4/4!......=2.154
x^4/24- x^2/2 + x =2.154-1
x^4/24- x^2/2 + x =1.154
by factorization; x[x^3/24- x/2 + 1] = 1.154
therefore; x= 1.154 or x^3/24- x/2 + 1 = 1.154

with this we have succeeded in getting on value of x=1.154, solve for other values

I disagree with u man, u choose the first three terms of d maclaurin's series, if my memory still dey, u can only make this assumption if u sure x<1, in dat case the rest of d series becomes insignificant. In this case u looking for the x, besides even ur factorization is kinda off. Ok, try feeding ur answer into the equation and c if it adds up.
Re: Nairaland Mathematics Clinic by 2good(m): 8:08pm On Jan 03, 2013
Adol16: Maths gurus pls help me solve the simultaneous equations

x + y =5
x^x + y^y=31

Graphical Solution is shown below.
We know that from first equation,
y=5-x which implies that y^y=(5-x)^(5-x)
Also from second equation,
y^y=31-x^x

So plotting a graph of y^y against x for both equations will give you 2 curve with a point of intersection, x=2 and y^y=27
y^y=27 implies that y=3

Re: Nairaland Mathematics Clinic by ositadima1(m): 8:11pm On Jan 03, 2013
^^^cool
Re: Nairaland Mathematics Clinic by Time2talk(m): 8:18pm On Jan 03, 2013
@ositadima1
Definitely
Re: Nairaland Mathematics Clinic by Richiez(m): 8:20pm On Jan 03, 2013
ositadima1:

I disagree with u man, u choose the first three terms of d maclaurin's series, if my memory still dey, u can only make this assumption if u sure x<1, in dat case the rest of d series becomes insignificant. In this case u looking for the x, besides even ur factorization is kinda off. Ok, try feeding ur answer into the equation and c if it adds up.

if you check the R.H.S, u'l find out that the value of x cannot exceed 2.154, and this is a small number compared to the next term which is to be divided by 6!, so this makes other terms quite insignificant.
and for the factorization, if you have an eqn, say, x(x^2+ 4)= 0 , one of the values of x must be 0. same goes for x[x^3/24- x/2 + 1] = 1.154, one of the values of x shud be 1.154
okay lets check
x+cosx=2.154
when x=1.154, x+cosx= 1.154 + cos(1.154) = 1.154 + 0.999 = 2.154 (approx)
Re: Nairaland Mathematics Clinic by dabrake(m): 8:20pm On Jan 03, 2013
ositadima1:

Guy dont stress ur self, I used a programming calc. and I got this answer;
a=4.87119
b=1.5644-2.51229i
c=1.5644-2.51229i

This is one set of answers, they are like 3 set total and they are all complex numbers, I doubt anybody can solve this equations by hand.

If a+b+c=8, a^2+b^2+c^2=16 and a^3+ b^3+c^3=64 find a,b, and c
chei! Omo, . . .

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