masperano: this oladoya way of solving maths is very hilarious (log cancels log) nairalanders ignore his solutions ooo lol very funny dude! @oladoya ur manipulations na wa ooo u na albert Einstein for manipulation

oga masperano if logn=logm, then n=m if think dat one is manipulation den try...
Log2^x=log4x
xlog2=log4+logx
xlog2-log4=logx
xlog2-2log2=logx
log2(x-2)=logx
if logn=logm
n=m
then 2(x-2)=x
2x-4=x
2x-x=4
x=4...
Oga better still you can solve it for us

mathefaro:
You don't seem to understand the laws of logarithms. Please, try to go and learn more about it cos the two solutions u gave contain lots of logarithm blunders. Hope i didnt sound too insolent

may be u should point the errors, so dat i can learn

oladoya: oga masperano if logn=logm, then n=m if think dat one is manipulation den try...
Log2^x=log4x
xlog2=log4+logx
xlog2-log4=logx
xlog2-2log2=logx
log2(x-2)=logx
if logn=logm
n=m
then 2(x-2)=x
2x-4=x
2x-x=4
x=4...
Oga better still you can solve it for us

xlog2 - 2log2 is not equal to log2(x-2) but (x-2)log2 which would in turn give log2^(x-2)

mathefaro:
xlog2 - 2log2 is not equal to log2(x-2) but (x-2)log2 which would in turn give log2^(x-2)

you see maths is all about using ur descretion. If you have 2x+2y you will have 2(x+y) den i only applied dat and lets stop the aegument you just solve ham

oladoya: you see maths is all about using ur descretion. If you have 2x+2y you will have 2(x+y) den i only applied dat and lets stop the aegument you just solve ham

lol! I wasn't and would never argue with you. I was only pointing out your errors. So get that. and for the solution, check the previous page, someone did it correctly by converting it to a quadratic equation and I also showed a more simple method of getting the result but if you want the logarithm version, here it is..

oladoya: bros this question is incorrect, it has to be in simultaneous form. Better still here comes the solving.....
2^x+3^x=13
introduce log.
Log2^x+3^x=log13
xlog2+xlog3=log13
x(log2+log3)=log13
xlog(2*3)=log13
xlog(6)=log13
x=log13/log6
log cancelled out and we have
x=13/6
x=2.16
app. 2

...thank you bro...it's just that you av already known the question that made you do it like that...moreso,in the logarithm,log(2^x + 3^x) can never be log2^x+log3^x

labodinho: I dey feel u ma guy,i've been longing to hit a thread like dis to express ma feeling bout maths....i luv yur workings thr on our thread...kan u check dah qustn on our thread n see if u can help cos i've tried n i'm stuck in d middle.

...could you please tell me the thread bro

ladokuntlad: pls help me find d value of x if x^2=4x. show workings oooo. hint x=4

...to do that...x^2=4x.then,x^2-4x=0...factorize x out to give x(x-4)=0...therefore x=0 or x-4=0.therefore x=0 or x=4

Pls.my oga at the top here...help me with this question...2^x=2x...add me on 2go with the username Mathemagician(if you know you are a maths guru) and let's gain from one another

x² = 4x;
take the natural log of both sides;
logx ² = log4x
2logx = log4 + log x;
collecting like terms;
2logx - log x = log 4;
logx = log4;
then if I may borrow your language;
"log cancels log";
x = 4.
U can't just use discretion in maths like you said because they're laws guiding everything you do. it's mathematics and not mathemagic.

Calculusf(x):
...thank you bro...it's just that you av already known the question that made you do it like that...moreso,in the logarithm,log(2^x + 3^x) can never be log2^x+log3^x

Exactly

ladokuntlad: pls help me find d value of x if x^2=4x. show workings oooo. hint x=4

x^2=4x..first apply log to both sides
logx^2=log(4*x)
2logx=log4+logx...by collecting lyk terms
2logx-logx=log4
logx=log4..log cancel log
::x=4

Mathefaro nd mr calculux have better understanding on the concept of lagarithmic function than oladoya! Oladoya get d principles clear man!but u re very hilarious sha hehehe (log cancels log)

Mr Calculus: x^2=4x..first apply log to both sides
logx^2=log(4*x)
2logx=log4+logx...by collecting lyk terms
2logx-logx=log4
logx=log4..log cancel log
::x=4

oga talk say make we no use dat word again, u no dey hear

oladoya: oga talk say make we no use dat word again, u no dey hear

Oladoya, pls u ae wrong........my boss mathefaro nd mr calculus ae right.
U cn still solve it dis way
x^2=4x, d.b.s by x
U will ave x=4
OR
x^2=4x
x^2 - 4x + 0=0
Factorise using general quadratic formula, u will have
x=4 or x=0
Therefore x=4.

Pls.help me on this integration...integral 7^sinx dx

Maths is fun for those who know it.

Maths is fun for those who knows it.

Boladearo: Maths is fun for those who knows it.

you're right.

oladoya: bros this question is incorrect, it has to be in simultaneous form. Better still here comes the solving.....
2^x+3^x=13
introduce log.
Log2^x+3^x=log13
xlog2+xlog3=log13
x(log2+log3)=log13
xlog(2*3)=log13
xlog(6)= log13
x=log13/log6
log cancelled out and we have
x=13/6
x=2.16
app. 2

It's against the law of logarithms for you to say log 13/ log 6 = 13/6. log a/log b is not= a/b. You can try this out if you don't mind.
(a) find x if 2^x = 6
(b) find x if log _x3 + log ₃x = 10/3.
they will help you understand the laws more. Please, no one is an island of knowledge, learning ends in the grave.

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

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.

Calculusf(x):
...could you please tell me the thread bro

Sir,itz Unilag admission thread 2013/2014.

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

.
.

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..

Calculusf(x):
Pls.help me on this integration...integral 7^sinx dx

integ7^sinxdx
Let u=sinx,du/dx=cosx,
dx=du/cosx,
integ7^sinx=integ7^udu/cosx,
recal integ a^x=a^x/Ina integ7^udu/cosx=1/cosxinteg7^udu=
1/cosx*7^u/In7+C
recal u=sinx..
1/cosx*7^sinx/In7+C=
7^sinx/cosxIn7 +C

Calculusf(x):
Pls.help me on this integration...integral 7^sinx dx

Let u=sinx
du/dx=cosx
dx=du/cosx
Therefore integral 7^sinx.dx=integreal 7^u du/cosx
1/cosx.integral 7^u.du
But integral a^x=a^x/lna
1/cosx(7^sinx/ln7)
=7^sinx/cosxln7+C

echibuzor:
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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...

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

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...

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

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.

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.

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

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.

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