freemandgenius:
Hey, where are the mathematicians here. Solve this with proves- 2^x=4x. Find x . Show full workings and earn a Christmas gift from me. The first to get it with proves.

freemandgenius:
Hey, where are the mathematicians here. Solve this with proves- 2^x=4x. Find x . Show full workings and earn a Christmas gift from me. The first to get it with proves.

freemandgenius:
Hey, where are the mathematicians here. Solve this with proves- 2^x=4x. Find x . Show full workings and earn a Christmas gift from me. The first to get it with proves.

kelvinreality:

guy, what about the Christmas Gift?

freemandgenius:
love that. You are 100% confident . Nice one. Will mail you soon for a token recharge. Drop your digits

freemandgenius:
love that. You are 100% confident . Nice one. Will mail you soon for a token recharge. Drop your digits

freemandgenius:
Hey, where are the mathematicians here. Solve this with proves- 2^x=4x. Find x . Show full workings and earn a Christmas gift from me. The first to get it with proves.

kelvinreality:


Christmas has passed and yet you did not fulfil your promise.

I hope you know Promise is a debt?

thankyouJesus:
Smiles

efficiencie:


your solution method is not a generalization as you will not arrive at the same answer if you were to change the interval [x n-x] where n=8...will you answer hold if the interval were [x n-x] where n<8 or n>8?

thankyouJesus:

Smiles....... You may need to read up the theorem sir.

Instead of going that length..... I will appreciate an example of an interval where it fails.

efficiencie:


Read my post well...I gave you a general set of intervals that will not provide you with a solution! I said will your answer hold if the interval were [x n-x] where n<8 or n>8?

thankyouJesus:
Picking of interval is not random as you may think..... I'm working based on intermediate theorem.. .... Let's assume I don't understand you...... Could you please give an interval to assert your point.

This is what I pick from your statement..... Let's assume n is 7..... I will be working with [4,7-4] = [3,4] right?

efficiencie:


Yes sir!

thankyouJesus:
Smiles..... Your interval wouldn't satisfy the mid value theorem I quoted because there does not exist a point that satisfies f(a) < 0 and f(b) > 0 totally.

efficiencie:


Sir just agree that your method is not a generalization. if a=3 and b=6 the condition f(3)<0 and f(6)>0 will be satisfied but the image of the mid-point f(4.5) is non-zero. Your solution begs the question; how did you determine the limits 3 and 4? Did you determine it by sheer guess work or by some mathematical reasoning you did not include in your solution?

efficiencie:


Sir just agree that your method is not a generalization. if a=3 and b=6 the condition f(3)<0 and f(6)>0 will be satisfied but the image of the mid-point f(4.5) is non-zero. Your solution begs the question; how did you determine the limits 3 and 4? Did you determine it by sheer guess work or by some mathematical reasoning you did not include in your solution?

thankyouJesus:
Smiles.... I really wish I can explain the theorem to you.. .. Check out mid value theorem and bisection method on any good Numerical Analysis textbook.. .. Shalom.. .

I need to do more iteration using your interval.. .. To your question.. .. It works for all.. .

Permit me to ask sir.. . Do you know and understand bisection method and mid value theorem?

efficiencie:


Sir, I am not challenging the veracity of the theorem I am merely showing you that your solution is not a generalization. Your choice of 3 and 5 as the bounds in your interval was either by mere inspection or sheer guess and not based on mathematical reasoning. Nevertheless I may be wrong so tell us how you arrived at the use of 3 and 5 as the bounds of the interval you used. In addition, your interval is not unique as there is an infinite set of intervals that will satisfy the theorem above and yield the solution x=4 for the problem f(x)=2^x - 4x where f(x)=0 [because by inspection, the interval [x n-x] where n=8 and x<4 will always yield your solution according to your stated theorem.]

thankyouJesus:
Smiles..... Check out any good Numerical Analysis textbook sir.. .. I ask again.. ..do you know and understand bisection method and mid value theorem sir?

freemandgenius:
Hey, where are the mathematicians here. Solve this with proves- 2^x=4x. Find x . Show full workings and earn a Christmas gift from me. The first to get it with proves.

kelvinreality:

2^x=4x, (apply log to base 10 to both side) log2^x=log(4x), xlog2=log4+logx, xlog2=log2^2+logx, xlog2=2log2+logx, (collect like terms), xlog2-2log2=logx, log2(x-2)=logx, (removin log to base from both sides ) 2(x-2)=x, 2x-4=x, 2x-x=4, x=4
x=4

Timology55:

Solu.
2^x =4x
Divide through by "4"
2^x/4 =4x/4
2^x/2^2=x
2^(x-2)=x......law of indices
The same as: (1+1)^(x-2)=x
Using Binomial theorem:
(1+a)^n=1+an+n(n-1)a^2/2! +......
Where a=1 and n=x-2
Substituting, we gat:
1+(x-2)+(x-2)(x-2-1)/2!=x
1+x-2 +(x-2)(x-3)/2=x
1+x-2 +(x^2-5x+6)/2=x
Multiply through by "2"
2+2x-4+x^2-5x+6)=2x
Simplifying the equation
x^2-5x +4=0.......quadratic equation
Factorize the quadratic equation
X^2-4x-x+4=0
(x^2 -4x)-(x-4)=0
x(x-4)-1(x-4)=0
(x-4)(x-1)=0
x=4 or 1
Checking: X=1 does satisfy the equation, hence it's an extraneous equation.
Hence, valid answer is x=4

X=4.....Q.E.D


My gift shouldn't be delayed

Evangkatsoulis:


your binomial expansion isn't complete. Why did you stop at the third term?

efficiencie:


Sir just agree that your method is not a generalization. if a=3 and b=6 the condition f(3)<0 and f(6)>0 will be satisfied but the image of the mid-point f(4.5) is non-zero. Your solution begs the question; how did you determine the limits 3 and 4? Did you determine it by sheer guess work or by some mathematical reasoning you did not include in your solution?

efficiencie:


sir, mathematics is a lovely discipline...I do not have to be a professor of mathematics to know that there is a flaw in your mathematical reasoning. My simple question to you is; HOW DID YOU DETERMINE THE INTERVAL [3 5]? WAS IT BY INSPECTION (Anyone could do that) OR BY SOME MATHEMATICAL REASONING (which you did not specify in your solution).

Timology55:


Solu.

2^x =4x

Divide through by "4"

2^x/4 =4x/4

2^x/2^2=x

2^(x-2)=x......law of indices

The same as: (1+1)^(x-2)=x

Using Binomial theorem:

(1+a)^n=1+an+n(n-1)a^2/2! +......

Where a=1 and n=x-2

Substituting, we gat:

1+(x-2)+(x-2)(x-2-1)/2!=x

1+x-2 +(x-2)(x-3)/2=x

1+x-2 +(x^2-5x+6)/2=x

Multiply through by "2"

2+2x-4+x^2-5x+6)=2x

Simplifying the equation

x^2-5x +4=0.......quadratic equation

Factorize the quadratic equation

X^2-4x-x+4=0

(x^2 -4x)-(x-4)=0

x(x-4)-1(x-4)=0

(x-4)(x-1)=0

x=4 or 1

Checking: X=1 does satisfy the equation, hence it's an extraneous equation.

Hence, valid answer is x=4


X=4.....Q.E.D



My gift