₦airaland Forum

fasodecapo: log₂(x² + 7x -2) = log₂(x² + 3x - 6) + log₄8;
log₂(x² + 7x -2) = log₂(x² + 3x - 6)+ 1/2log₂8;
log₂(x² + 7x -2) = log₂(x² + 3x - 6) + log₂8^3/2;
log₂(x² + 7x -2) = log₂(8^3/2)(x² + 3x - 6);
x² + 7x -2 = (8^3/2)(x² + 3x - 6);
please check the question very well if it's truly base 4, if it is notify me to finish this.

U are stuck at find a root for 8 abi? same here.. lets see if someone else have got ideas...

labodinho: Echibuzor.....Boss plz chk dis.

labodinho: Log(x sq + 7x - 2)
base 2 equal to Log(x sq +
3x -6) base 2 + Log8 base 4

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.I can see the answer in my head... I am on my way to work.. Solution loading...

labodinho: sir,i got the mail.Tanx for the help.The answers were not in d options buh if datz d procedure for working ,there's no probs.No 1 optionz (A)6(1/3)(B)5(1/2)(C)9 (D)5(2/3).....No. 2 (A)2 (B)1(1/3)(C)1(1/99)(D)No solution....No. 3....No options

The option is there now.. 1. is D... 2. B... Did.u bother to look for typos in the document... read and study it...

oladoya: here comes the answer... (1) log9-log3+log9 since they are in diff base just represent it with letter so we have A-B-C so lets solve A... Log9base3=a, then let the base have a as its power then we have 9=3^a, 3^2=3^a equate the base we have a=2.... Then lets take B... Log3base27=b, 3=27^b, 3^1=3^3b equ base, we have b=1/3.... Now C..... Log9basert3=c, 9= 3^c/2 we have 3^2=3^c/3 equ base we have 2=c/2 cross mult. We c=4 then remenber A-B+C = 2-1/3+4=17/3.... 2nd question (logx)base2-(logx-1)base2=2... According to the rule take a log(x/(x-1))=2, x/(x-1)=2^2, x/(x-1)=4 cross multiply you have x=4x-4, x-4x=4, -3x=-4, x=4/3...... No3 solving it quadratically we have x>1/2 or x>-4

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. Good man. but the problem here i that if I dont a good knowledge of the subject, it will look like jibberish to me....
I think Seun has to upgrade the Text Editor to one that can handle complex typefaces...

labodinho: Tanx bro.

Check your Email box...

labodinho: Oh,tanx.can't really see them clearly.Can u mail me on labodihoo@gmail.com or exercise patient by writing it.Plz,jst @ yur conveniency....

Alrite man.. I ll type it on Word and attach it to the mail I am gonna send gonna send to u..

labodinho: Plz help me out.Hw do i solve logs with diff base like this No 1 Log9 base 3-Log3 base 27+ Log9 base sqaure root3.........No 2.Solve the equatn LogX base 2-Log(X-1) base 2=2.........This one is on Inequalities.No 3. (2x-1)(x+4) divided by x-5 both numerator n denominator Greater than(sign) 0 ....Tanx.

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I couldnt type it but I hope the picture of my workbook is clear enough...

I tried not to laugh but some of these comments are damned funny..
@OP... Forget it, the best u can do with #1,000 is to spend it on food or Recharge Voucher...

annoymous: How much is a bucket of alomo?

BRAINEE123: Plz pals , solve this for me: If a body moves with a final velocity of 30km/hr and accelerates at 12km/hr^2 in 2hrs. What is the distance covered in 2.5hrs?

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. This is a problem under the Laws of motion..
The first part of the question..(a body moves with a final velocity of 30km/hr and accelerates at 12km/hr^2 in 2hrs)...
V = u + at, where V = 30, a = 12, t = 2, by substitution, u = 6.....
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The second part of the problem (What is the distance covered in 2.5hrs?)...
S = ut + 1/2 at².... where u = 6, a = 12, t = 2.5..
by substitution, S = 36km...

Edis4christ: No chieef, its nt correct dat way, u will ave 2 follow d laws of logarithm

Thanks man.. My Logarithms is rusty...
Learnt something new..

Boladearo: (81^1/log3) / (36^1/log6). The logs are in base 2.

I am assuming that the question says 81^(1/log3) / 36^(1/log6)
.From laws of logarithn, 1/logb base a=loga base b
So we ave 81^log2 base 3/ 36^log2 base 6
3^4log2 base3/ 6^2log2 base 6
But b^ylogx bseb=ylogx baseb=logx^y baseb
3^log2^4 bsae3/6^log2^2 base6
4rm laws of log a^logb base a=b
2^4/2^2=16/4
=4.
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. Courtesy of General Edis

Boladearo: (81^1/log3) / (36^1/log6). The logs are in base 2.

Is it like this: 81^1/(log3) / 36^1/(log6) or like this : 81^(1/log3) / 36^(1/log6) ?

Boladearo: 81^1/log3 divide by 36^1/log6, the logs are in base 2.

Please use brackets to clarify the expression.. Thank You..

masperano: Donedy dere is no problem in wat ortarico has done we all know the expresion is invalid for y=2! Abi u b cauchy wey introduce rigor into maths?dats y I luv euler! My all time best mathemtician

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General Masperano dont be so sure about that conclusion that everyone knows.

donedy: Your calculation is right, per say. But next time try to state the condition under which the solution is valid. You cannot just divide both sides of the equation by an algebraic expression without some conditions.

Basically, when you divided both sides 2y - 4, you should have explicitly stated "for y != 2", for deduction to be valid.

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Thats a good observation Sir, So as not to give an undefined solution at y=2.

Ortarico: nice one @echibuzor. . . .thanks for the hint
Here it is:

y= 4x + 3/2x - 5
cross multiplyng gives:
y(2x - 5) = 4x + 3
2xy - 5y = 4x + 3
collect like terms:
2xy - 4x = 3 + 5y
x(2y - 4) = 3 + 5y
dividng both sides by 2y - 4 gives:
x = 3 + 5y/2y - 4
re-arrangng gives:
x= 5y + 3/2y - 4. . . . . .answer

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Good Man... You are welcome, I was reading the question as y = 4x + (3/2x)- 5, it shows what a difference a pair of brackets can make in mathematics...

labodinho: yea,i jst got dah,kan u n @ all check on these....(1)solve the equation _/x+7=x-5 (2).....Let Y=4x+3/2x-5,write x as a function of y. (3).....If dy/dx=6xrtp2 + 15xrtp4 and y=7 when x=2.Find y N.B:rtp=rays to power....tanx

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labodinho: Helo ma General's,i'm sorry for not
posting since i posted this
question.I've not been able to get
on line,very busy sir's.Tanx for d
working...No 2 qustn still pending
sir's.Tanx for the help.

When u hear in maths that A is a function of b then it means that A is the subject of the formular... , I hope that is a major hint to solve no 2.. Workings loading, I am mobile at the moment...
y = 4x + (3/2x)- 5.. Wait a minute, or is the question y = 4x + 3/(2x- 5) ? Please specify..

Mr Calculus: plz guyz need ur help:::Convert 1111.100110 base 2 to base
12 and 123.X4 base 12 to base 2 without converting first to base 10

I have an idea... Depends on the method of coversion u use.. If u use the 124816 method, then u should find the equivalents of each bits in base 12 then use it as ur subscripts... Am I making any sense to u?

biolabee: This is the 1 - P (2 jobs) or P (no jobs)

P(2 jobs) = 0.2 *0.2 = 0.04
P(no jobs) = 0.8 * 0.8 = 0.64

1 - 0.64-0.04 = 1- 0.68 = 0.32

You can also do P(one job, no job = 0.8 * 0.2 = 0.16
Multiply by 2 = 0.32

Possiblity

P (JJ), P(JN), P(NJ)and P(NN)

Good man, I was reading the question as dependent probability... I need to see my optician ASAP.. anyway, good job..

sniperwolf: A manufacturing company is independently working on two
separate jobs. If the probability that either of the jobs will be
finished on time is 0.02, the probability that just one of the jobs
will be finished on time will be ....
[ICAN May 2013. Costing and Quantitative Techniques]

Is this Question complete? before i even attempt...

knockturnal: Its a way of determining if the victim is awake or sleeping. if u dont respond by slapping at them, then they know its safe to feast on your iron rich blood!
thats all!

Ur Signature is hilarious man.. Perfect word gan ni

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I dont know the section this is supposed to go to but I decided on dropping it here.
This application for employment was mailed in response to the advert we put out. I botched the name, address, phone number, and organization so that i wont be sued or anything like that..

WHξξ∟s:
Hello, as the saying goes, understanding a question is part of answering the question, but the unfortunate thing is what i have here is something that am not familiar with. I dnt seek for answers only i also like to know how i shall work it out on my own and provide solutions[dont just give me the fish, show me how to catch it also].

Question 1;

[img]http://f12.wapkafiles.com/download/d/0/e/297960_d0efe68cb55d2f5f2f7188ea.PNG/aeb99fdd70705acd7939/6.PNG[/img]

Question 2;

[img]http://f12.wapkafiles.com/download/7/a/f/297960_7affa0a3c7c6bc5ed401c4c5.PNG/fcc91f7c615fd4deea03/8.PNG[/img]

Question 3;

[img]http://f12.wapkafiles.com/download/3/f/d/297960_3fd30711fa59d6d21390969b.PNG/46a8d85f7dcfd0825b2c/9.PNG[/img]

Sorry, but I cant see the picture... i am mobile... try transcribing it...

I hope they brought that Trophy to d stadium

Van Persie 1 - Gerrard 0

johnpaul1101: Please someone should help me out with this: solve cos345 without the use of calculator or table

That's Cos 15 in the fourth quadrant.... Solution Loading


cos 2x = 2cos^2 15 - 1
2cos^2x = cos2x + 1 ... x = 15

2cos^2 (15) = cos 30 + 1 .... Cos 30 = (√3)/2

2cos^2 (15) = [(√3)/2] + 1

cos^2 (15) = ([(√3)/2] + 1) /2

2cos^2 15 = [(√3) + 2] /2

cos^2 15 = [(√3) + 2] /4

Square root both sides...
cos15 = ±√([(√3) + 2] /4)

cos15 = (√[(√3) + 2])/2

and that, my friend, is cos 345 = cos 15 = (√[(√3) + 2])/2...

Did u notice that i removed the ± in the final answer? That's because Cos is always positive in the fourth quadrant....

At first I was solving for sin15 before i looked at the question again... But, Hey, U might need it too someday right?

So, Here goes, Its quite lengthy and questions about further explanations are welcomed... Lets go there....
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Sin (2y) = 2sinycosy ------ (by putting y=15)
Sin30 = 2sin15cos15 ------(ii) (cos^2 (y) = 1 – sin^2 (y))
Hence, cosy = √(1 – sin^ (y))
By putting this into eq (ii) we have;
Sin30 =2sin15 * √1-sin^2 (15)
Square both sides…
Sin^2 (30) = 4sin^2 (15) (1 – Sin^2 (15))
Open Bracket to give…
Sin^2 (30) = 4sin^2 (15) – 4sin^4 (15)
Remember that Sin ^2 (30) = Sin30 * Sin30 and Sin 30 = ½ (abi?)
So: ½ * ½ = 4Sin^2 (15) - 4Sin^4 (15)
¼ = 4Sin^2 (15) - 4Sin^4 (15)
Cross Multiply…..
1 = 16Sin^2 (15) – 16Sin^4 (15) =>>> 16Sin^4 (15) -16Sin^2 (15) +1 = 0
What we have here is a quadratic equation of the Sin^2 form
Let Sin^2 (15) = x
Our new equation becomes…..
16x2 – 16x + 1 = 0
Using Quadratic Formulae (Popularly known as Almighty Fomular)
x = [-(-16) ± √162 - 64 ] /32
==>> [16±√256 – 64] /32
==>> [16 ± √192] /32 ==>> [16 ± √64 * √3] / 32
==>> [16 ± 8√3] /32 …… By Simplifying with a common factor (
==>> [2 ± √3] /4 ]
Sin^2 (15) = [2 ± √3] /4]
Sqrt both sides….
Sin15 = ± (√[2± √3] /4 )
This is where the logic comes in… Remember that I said Sin345 is Sin 15 in the fourth Quadrant (Fourth Quadrant is angles between 2710 and 3600) and any Sine in the Fourth Quadrant gives a negative answer… This point eliminates + √[2± √3] /4 from being a valid answer, which gives us with -√[2± √3] /4… Also (√2+√3) / 4 gives a value too extreme to be relevant… Hence, we are left with
-√[2 - √3] /4 as our answer for sin15
To have a more simplified answer… -√[2 - √3] /4 is also the same as [√(2 - √3)] / √4 which gives -[√(2 - √3)] /2 …….. That is your answer… Confirm with a calculator… Questionand request on further explanations are welcome…

Chibuzor..

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That Fugly Woman has her lips upside down...the brain rotate it when looking at it at first but leaves it alone when u turn d phone/whatever u are viewing with...

2nioshine: Really impresed wsh i could remain always on d tread lyk wen it was newly opend............bt sku strez keeps........@Ric,echibuzor,biola,gh.guru and all who az contributd 'RESPECT'....Where iz my man Double dx @?

U too man.. u too

Richiez: who can resolve this into partial fractions
(3x + 2)/(x² + 1)²

Loading......... couldnt arrive at a valid answer.