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

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

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.

Thanks but I thought all would understand. Most of us aren't Learners!

Edis4christ:
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.

yea you a right, the question is x^2=4x but i misquote it for 2^x=4x. Anyway thanks guys

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

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) ?

This the question, in image format

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

If its 81^1/(log3)/ 36^1(log6) all in base 2 den it will b
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

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.
.
. Courtesy of General Edis

Edis4christ:
If its 81^1/(log3)/ 36^1(log6) all in base 2 den it will b
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

nyc work bros. It is correct

echibuzor:
.
.


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..

itz y=4x+3 divided by 2x-5

echibuzor:

I am assuming that the question says 81^(1/log3) / 36^(1/log6)
.
next step is divide the 81 by 36 then subtract the powers...
.
(81/36)^{1/log3 - 1/log6}

(9/4)^{1/log3 - 1/log6}
.
using LCM for the Powers,,, (log6 - log3) /(log3.log6).........(log6 = log(2*3) = log2 + log3)
.
(log2 + log3 - log3) / log3.log6
(Log3 cancel Log3 {LOL}) and the question says the logs are of base 2 therefore log2 = 1
.
That gives the power as 1/log3 (log3 + log2)

= 1/log3 (log3 + 1)

The whole solution now becomes...
(9/4)^{1/log3 (log3 + 1)}

Thats the point I could go no further.. Additions are welcome.

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

Edis4christ:
If its 81^1/(log3)/ 36^1(log6) all in base 2 den it will b
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

I tink 4 is d correct ans, 4rm my solvings, additions ae welcome though

Edis4christ:
I tink 4 is d correct ans, 4rm my solvings, additions ae welcome though

it is correct bros, but any other solution is welcome sha

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

NOTE::¥=divide
Recall=logxbasey=1/logybasex,
81^1/log3¥36^1/log6=81^log2base 3¥36^log2base6,
81=3^4,36=6^2.
3^4log2base3¥6^2log2base6,
3^log2^4base3¥6^log2^2base6,
3^log16base3¥6^log4base6,
RECALL=a^logxbasea=x,
3^log16base3=16 and 6^log4base6=4
ur ans=16/4=4

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..

Ortarico:

Greetings General doubleDx.
No qualms my email is oketundericharda@yahoo.com. What's yours and are you on facebook? It's good to have someone like you on my friend's list.

Aight man, I'm on mobile now, will email you later 2day! 1luv.

9ice work guys, keep up the good work!

doubleDx:

Aight man, I'm on mobile now, will email you later 2day! 1luv.

9ice work guys, keep up the good work!

All right!

Mr Calculus:
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

...i understand this bro but it's not the answer coz you know u=sinx therefore du/dx=cosx...and dx=du/cosx...so the question becomes integral 7^u.du/cosx...you just took the cosx out of the integral function as if it's a constant...so,it can't be that bro...how can one integrate it from there...

[color=#990000][/color]pls permit me to tease your brain with this primary school maths... 1.prove that (a+b)^2=a^2+2ab+b^2 2.solve (a-b)^2

Megaekevisky: [color=#990000][/color]pls permit me to tease your brain with this primary school maths... 1.prove that (a+b)^2=a^2+2ab+b^2 2.solve (a-b)^2

Replace b in the first equ with -b

(A+ -b ) ^2 = a2+2 a x-b + (-b)^2
= a2-2ab+b2

u are wrong man

[quote author=biolabee]

u are wrong, pls resolve

Guys imagine this




A Texas banker is upping the ante to $1
million for whoever solves a tricky problem
that's been dogging mathematicians since
the 1980s.
The Providence, Rhode Island-based
American Mathematical Society (AMS) on
Tuesday said $1 million will be awarded for
the publication of a solution to the Beal
Conjecture number theory problem.
Dallas banker D. Andrew Beal first offered
the Beal Prize in 1997 for $5,000. Over the
years, the amount has grown.
WHO'S GOT WHAT IT TAKES TO SOLVE THE
PUZZLING 'BEAL CONJECTURE?
If Ax + By = Cz , where A, B, C, x, y and z
are positive integers and x, y and z are all
greater than 2, then A, B and C must have a
common prime factor.




https://www.nairaland.com/1314580/banker-offers-1million-prize-anyone

Megaekevisky: xx

not sure but will look at it again

Calculusf(x):
...i understand this bro but it's not the answer coz you know u=sinx therefore du/dx=cosx...and dx=du/cosx...so the question becomes integral 7^u.du/cosx...you just took the cosx out of the integral function as if it's a constant...so,it can't be that bro...how can one integrate it from there...

He didnt take dat du/cosx out like dat. dx=du/cosx
Imtegral 7^u . du/cosx is dsame as
(1/cosx)integral 7^u . du.

Calculusf(x):
...i understand this bro but it's not the answer coz you know u=sinx therefore du/dx=cosx...and dx=du/cosx...so the question becomes integral 7^u.du/cosx...you just took the cosx out of the integral function as if it's a constant...so,it can't be that bro...how can one integrate it from there...

Bro in integratn if u av
integ2x it is dsam as 2integx.
if u av integx/2 it is dsam as 1/2integx.
u can try solving d exampls above u'l 5nd out dat its dsam.
I hope u undstand

I really need ur help hear guys.
If a+b+c=0,simplify::
[{x^(a^2 b^-1 c^-1)}]*[{x^(a^-1 b^2 c-1)}]*[{x^(a^-1 b^-1 c^2)}]
Given that::a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)
NOTE: * means multiply

RICHIEZ THIS IS YOUR OPPORTUNITY TO BECOME A MILLIONAIRE.



A Texas banker is upping the ante to $1 million for whoever solves a tricky problem that’s been dogging mathematicians since the 1980s.

The Providence, Rhode Island-based American Mathematical Society (AMS) on Tuesday said $1 million will be awarded for the publication of a solution to the Beal Conjecture number theory problem.

Dallas banker D. Andrew Beal first offered the Beal Prize in 1997 for $5,000. Over the years, the amount has grown.

AMS spokesman Michael Breen says a solution is more difficult than the one for a related problem – Fermat’s Last Theorem – which didn’t have a published solution for hundreds of years.

Mr Beal is a self-taught mathematician and founder of the Beal Prize. He says he wants to inspire young people to pursue math and science.

An AMS-appointed committee will award this prize for either a proof of, or a counterexample to, the Beal Conjecture, published in a refereed and respected mathematics publication.

The prize money is being held in trust by the AMS until it is awarded.
Income from the prize fund is used to support the annual Erdős Memorial Lecture and other activities of the Society.

According to the AMS website, the $1million prize will be awarded to anyone who can solve and fulfill the following:

If Ax + By = Cz , where A, B, C, x, y and z are positive integers and x, y and z are all greater than 2, then A, B and C must have a common prime factor.


The administration of the Beal Prize is overseen by a Beal Prize Committee (BPC) to be appointed by the President of the AMS.

The formal charge of the BPC and the ‘Procedures for Determination of an Award of the Beal Prize’ are subject to the review and approval by the Council of the AMS.

The Beal Prize Fund is held as a restricted asset of the American Mathematical Society (AMS), with US$1,000,000 to be awarded if, in the judgment of the BPC, the conjecture is proved or a counterexample is presented.

A proposed solution of the Beal Prize Problem may not be submitted directly to the AMS, or to the Beal Prize Committee, or to Mr Beal. Unpublished manuscripts will not be considered.

The BPC will consider a proposed solution if it is a complete mathematical solution of the Beal Prize Problem. Before consideration, a proposed solution (the ‘Work’) must be published in a refereed mathematics publication which is respected and, in the opinion of the BPC, maintains the highest editorial standards (or published in another form as the BPC decides may qualify).

In the case of a counterexample, the proposed solution will be subject to independent verification. Upon publication, the author(s) of the Work should notify the AMS and the BPC.

The Work must be widely accepted by the mathematics community following a two-year waiting period after publication.

In the case of a counterexample, that recognition and acceptance by the community may happen much sooner.

Following the waiting period, the BPC will decide whether the Work merits detailed evaluation.
If the Work is to receive detailed evaluation, the BPC and the AMS will identify at least two experts who can verify the correctness of the Work and who are not members of the BPC to assist in the evaluation.

Upon completion of the evaluation, if the BPC can make a clear decision, it may award the prize and determine attribution of credit for a solution.

If Ax + By = Cz , where A, B, C, x, y and z are positive integers and x, y and z are all greater than 2, then A, B and C must have a common prime factor.

Mr Calculus: Bro in integratn if u av
integ2x it is dsam as 2integx.
if u av integx/2 it is dsam as 1/2integx.
u can try solving d exampls above u'l 5nd out dat its dsam.
I hope u undstand

.i know that...but you as mr calculus said integral 7^u du/cosx it's not possible to take cosx out just as if it's a constant.hadn't been it's ln2 or 7 or any constant...it can...but this one can not be done like that...THOUGH...i don't know the question but if someone gets it i will know

Hey guys. Pls I need a quick method to find d drterminant of a 4x4 matrix and ti find the eigen values
using cayley hamilton, expect frm d normal way cos its too long... Urgent reply pls

Lumidey: Hey guys. Pls I need a quick method to find d drterminant of a 4x4 matrix and ti find the eigen values
using cayley hamilton, expect frm d normal way cos its too long... Urgent reply pls

...try sarrus method bro...