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Differential Calculus

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24 Feb 2008 11:53:32 IST

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453

Exponential eqn

None

Find x such that

2001^x+2004^x = 2002^x+2003^x

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Comments (9)

anchit saini

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24 Feb 2008 12:19:23 IST

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Akhil

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24 Feb 2008 12:19:26 IST

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2002^x-2001^x=2004^x-2003^x
2002=2001+1

(2001+1)^x-2001^x=(2003+1)^x-2003^x

expanding we get

1+^xC₁(2001)+^xC₂(2001)²+....^xC_x-1(2001)^x-1=1+^xC₁(2003)+^xC₂(2003)²+....^xC_x-1(2003)^{x-1

so only value for x can be 0 and 1}

Hari Shankar

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24 Feb 2008 12:29:28 IST

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Good. Any other approaches. You can take a hint from the category under which it is posted.

I said any other approach, because the above one is not completely legitimate. I dont think all of us are comfortable with binomial theorem applied to real numbers. There is something more to be done

sti I

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24 Feb 2008 20:29:10 IST

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2001^x+2004^x/2002^x+2003^x=1

takin log on both sides

log 2001^x+2004^x=log 2004^x+2003^x

differentiating on both sides and using formula

d/dx(a^x)=a^x log a

d/dx(log x)=1/x

we will get the ans

hope method is right!!

Hari Shankar

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26 Feb 2008 13:33:18 IST

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Method I:

Write the equation as 2004^x - 2003^x = 2002^x - 2001^x

Now, consider the function f(x) = p^x - (p-1)^x for p>1.

f'(p) = x(p^x-1 - (p-1)^x-1)

You can see that f'(x) = 0 for x=0 or x=1.

for any other value of x, f'(p) is a strictly monotonic function.

which means 2004^x - 2003^x = 2002^x - 2001^x only if x=0 or x =1.

Abhijith

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26 Feb 2008 13:53:19 IST

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To prove:

CASE 1 : x > 1

It can be easily shown that

and as

Hence no solution for x > 1.

CASE 2: x < 0

Let

sucht that

(proved above) and

Thus there are no solutions for x < 0 .

Case 3: x = 0 or x = 1

These 2 are the only solutions.

edit: I was typing out my solution when you posted your method, i guess this one is not so different.

Hari Shankar

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26 Feb 2008 14:19:38 IST

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Koni: What happens for 0<x<1.

Abhijith

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26 Feb 2008 19:13:43 IST

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Similary,

So the graphs of both f(x) and g(x) are concave upwards (a parabola opening upwards). Any 2 functions which satisfy this condition can intersect only at 2 points.

Hari Shankar

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26 Feb 2008 19:52:47 IST

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There is one more approach, and this line of reasoning could also prove useful:

Again rewrite as 2004^x - 2003^x = 2002^x - 2001^x

Now, consider f(t) = t^x.

By Intermediate Value Theorem, there exists p

[2004,2003] such that

xp^x-1 = 2004^x - 2003^x /(2004-2003) = 2004^x - 2003^x

Similarly there exists q

[2004,2003] such that

xq^x-1 = 2002^x - 2001^x /(2002-2001) = 2002^x - 2001^x

Hence xp^x-1 = xq^x-1

or x(p^x-1 -q^x-1) = 0

Since p and q are distinct, x=0 or x=1 are the only possibilities.

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