Neyozak:

So far so good, this is the best approach. Invitation loading.

Well if I may pinpoint some error here
How can a variable equals to itself
>>> x=2^(2x-3)........eqn(2)
Is Dat mathematical ?
I stand to be corrected Dat is an error , I don't mind if d OP CAN EXPLAIN TO ME BETTER

I am 100% correct with my solution. I am open to criticism@Neyozak

u try small but how 2x^2 turn to 4x^2

flossy007:
2^ 2x = 8x

Log100 ^2x = 8x ( log to base 10)

100^2x = 10^8x

(10^2) ^ 2x = 10^8x

(2x)^2 = 8x

4x^2 = 8x

X^2 = 2x

X = 2

Hope I got it

This ans is almost perfect so op no ojoro

Adehorze:
2^2x=8x
Taking log of both sides,
Log2^2x=Log8x
2xLog2=Log8+Logx
2xLog2=Log2^3+Logx
2xLog2=3Log2+Logx
2xLog2-3Log2=Logx
(2x-3)Log2=Logx
Equating d log of both sides,
>>
(2x-3)2=x
4x-6=x
4x-x=6
3x=6
Dividing both sides by 3
3x/3=6/3
x=2

Ur manipulation to arive at eqn 2 is not clear and may not be mathematical. How can the expression 2^2x = (2^3)^3 be divided by 2^3 to yield 2^(2x-3)?
provide the proof

chiv:
Just got the answer
2^2x=8x
2^2(x)=8x
4^x=8x........eqn(1)
2^2x=8x
2^2x=(2^3)x
Divide both sides by 2^3
x=2^(2x-3)........eqn(2)
Substitute 2^2(2x-3) for x in eqn(1)
4^(2^(2x-3))=8(2^(2x-3))
16^(2x-3)=2^3(2^(2x-3))
2^4(2x-3)=2^3(2^(2x-3))
2^4(2x-3)=2^3(2^2x)/2^3
2^4(2x-3)=2^2x
Equate the powers of 2 on both sides
4(2x-3)=2x
8x-12=2x
-12=2x-8x
-12=-6x
x=2

That's (2x)^2 i.e 2x multiplied by 2x

prideofscience:
u try small but how 2x^2 turn to 4x^2

Adehorze:


Well if I may pinpoint some error here
How can a variable equals to itself
>>> x=2^(2x-3)........eqn(2)
Is Dat mathematical ?
I stand to be corrected Dat is an error , I don't mind if d OP CAN EXPLAIN TO ME BETTER

Believe me without any sentiment the guy got it. That problem is just to exercise the brain to think out of box. Am was quite sure dat somebody will get d steps but am supprised that the solution steps came earlier than expected. I though it may probably take us few days to av it done. U see what dat guy did was just logical manipulation which satisfied the mathematical statement or equation. Test his manipulation for x equal to 2. U will see it is defined. On my own marking guide I divided by 8 with virtually similar steps. Let me shock u, at times we add zero to equation to be solved e.g +1-1,et.c All the same thanks.

Godiskind:
Ur manipulation to arive at eqn 2 is not clear and may not be mathematical. How can the expression 2^2x = (2^3)x be divided by 2^3 to yield 2^(2x-3)?
provide the proof

Check well it is simple indices rule! Meanwhile ur post was corrected, u had error on ur post.

IF YOU CANT PROOF IT, WE WILL REMOVE THIS POST AND BAN U FOR 6MONTHS FOR UNBASED MATHEMATICAL FACT

Neyozak:


Check well it is simple indices rule! Meanwhile ur post was corrected, u had error on ur post.

THANKS TO ALL DAT VIEWED AND CONTRIBUTED. Ese adupe and I hope someone had learnt one or two things.

This argument must go to front page for resolutions

Cc: Lalasticlala

[quote author=Neyozak post=38757493]
Believe me without any sentiment the guy got it. That problem is just to exercise the brain to think out of box. Am was quite sure dat somebody will get d steps but am supprised that the solution steps came earlier than expected. I though it may probably take us few days to av it done. U see what dat guy did was just logical manipulation which satisfied the mathematical statement or equation. Test his manipulation for x equal to 2. U will see it is defined. On my own marking guide I divided by 8 with virtually similar steps. Let me shock u, at times we add zero to equation to be solved e.g +1-1,et.c All the same thanks.
[/quote

Guess u don't understand my point of view
Just pay attention

x=2^(2x-3) .....eqn 2
Wat d above equation means is DAT u will substitute for x in eqn 1

>>> 4^x = 8x
So
4^2(2x-3) = 8(2^{2x-3})
From here u will have to substitute for x again which meke it to be expanding

NOTE

How can x= 2^(2x-3)
X cannot equals itself .

[quote author=Adehorze post=38758550][/quote]


It is possible my brother. With that alone check for x equal to 2 u see it is defined.
Even x equal x in the original question. learn my brother. Dat mathematics u can not no all. I know 4 sure dat someone's peak is another person's beginning. Learn bro!!!

Karmanaut:
Via numerical method (The Lambert W function to be precise)
There are two solutions to this equation:
x=0.154953, x=2.

even though you are correct, ironically, u were ignored.
no other solution posted here is correct

Suppose u have a question which is if
2^x =2x, find x?
SOLUTION
Any equation like dis usually involves logarithm to solve, I'm not doing show off or over sabi, just sharing my knowledge .

2^x =2x
Taking log of both sides
>> log2^x =log2x
xlog2 = log2 + Logx
Collecting d like terms
xlog2-log2 =Logx
Since log2 is common,
>>>
{x-1}log2 = logx
From above since d logs are in dsame level, so we equate d logs
>>
{x-1}2 =x
2x-2=x
Collecting like terms
2x-x=2
x=2

I no grey my ans = -1, is correct

those ur step is not acceptable in matimatical world

.

Neyozak:


U just manipulated, however u tried.

u need a reasonable and mathematical acceptable (dat is, with maths laws) manipulation to solve a 'hard nut to crack' lyk dis, guy. We hv many genius here.
He is correct

Adehorze:
Suppose u have a question which is if
2^x =2x, find x?
SOLUTION
Any equation like dis usually involves logarithm to solve, I'm not doing show off or over sabi, just sharing my knowledge .

2^x =2x
Taking log of both sides
>> log2^x =log2x correct
xlog2 = log2 + Logx correct
Collecting d like terms
xlog2-log2 =Logx correct
Since log2 is common,
>>>
{x-1}log2 = logx correct
From above since d logs are in dsame level, so we equate d logs
>>
{x-1}2 =x wrong
2x-2=x
Collecting like terms
2x-x=2
x=2

Dis is d only condition where u can equates logs of both sides

log 2^(x-1)= log x

I wanna thank u all

I won't argue wit u again but I've a cracker for u since u are my boss cos I've been unable to solve it.

If
x+y=5.....eqn1
x^x+y^y=31....eqn2
Find x&y

THIS CAN ONLY BE SOLVED NUMERICALLY...WAS ONCE OBSESSED WITH SUCH TYPE OF QUESTION IN SENIOR SCHOOL.@OP IF I SOLVE AM HOW MUCH YOU GO PAY ME?

hilaomo:
THIS CAN ONLY BE SOLVED NUMERICALLY...WAS ONCE OBSESSED WITH SUCH TYPE OF QUESTION IN SENIOR SCHOOL.@OP IF I SOLVE AM HOW MUCH YOU GO PAY ME?

I've solved it.
Check page 1.

Karmanaut:
Via numerical method (The Lambert W function to be precise)
There are two solutions to this equation:
x=0.154953, x=2.

Ok

Ok let let us continue mathematicians in d house !

Adehorze:

If x+y=5.....eqn1 x^x+y^y=31....eqn2 Find x&y

Solutions to these equations will highly be appreciated.

Adehorze:

If x+y=5.....eqn1 x^x+y^y=31....eqn2 Find x&y

Solutions to these equations will highly be appreciated.

Neyozak:

Solutions to these equations will highly be appreciated.

I'm expecting u to solve it for me Boss.

Neyozak:

Solutions to these equations will highly be appreciated.

Solution:
x=3, y=2
or
x=2,y=3.

PS, for these questions, Nairaland has a mathematics clinic

Simple solution. X=0. Details below

2^2x=8x

expand 2^2x will be 4x(square)

4x(square) =8x

divide both side by 4x

4x(square)/4x=8x/4x

x=2

Very simple!
2^2x = 8x
Multiply both sides by log base of 10
Log2^2x = Log8x
2xLog2 = Log8 + Logx
2xLog2 = Log2^3 + Logx
2xLog2 = 3Log2 + Logx
Therefore,
2xLog2 - 3Log2 =Logx
(2x-3)Log2 = Logx
By comparison (L.H.S & R.H.S)
2x-3 =1
2x =3+1=4
x= 2
Or a straight forward approach
2=x

Who will dispute this? @Neyozak, where is the Consultancy offer?