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Algebra

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Joined:	25 Jun 2007
Post:	14

17 Aug 2007 19:21:35 IST

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Experts, help me out, plz

None

Show that,

5 ^x + 2 ^x = 4 ^x + 3 ^x

has only 2 real solutions.

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Titun

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Joined: 23 Dec 2006

Posts: 374

17 Aug 2007 19:45:49 IST

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Consider a function,

f (t) = t ^x

Here, f (5) = 5 ^x, f (4) = 4 ^x , f (3) = 3 ^x , f (2) = 2 ^x

Again, f ' (x) = x.t^{x - 1}

The above function is continuous and differentiable throughout its domain.

Now, before I proceed, let me tell you that I will be using Lagrange's Mean value theorem to solve the problem.

Lagrange's mean value theorem for a function g (z) which is defined, continuous and differentiable througout the closed interval [ a, b ], there exists a point 'c' in the open interval of ( a,b ) where,

f ' (c) = [ f (b) - f (a) ] / (b - a)

Now, 5 ^x + 2 ^x= 4 ^x + 3 ^x

5 ^x- 4 ^x = 3 ^x - 2^x

f (5) - f (4) = f (3) - f (2)

[ f (5) - f (4) ] / (5 - 4) = [ f (3) - f (2) ] / (3 - 2)

f ' (c) = f ' (d) [ by Lagranges theorem, where 4 < c < 5 & 2 < d < 3 ]

x. c ^{x - 1}= x . d ^{x - 1}

x [ c ^{x - 1} - d ^{x - 1} ] = 0

x = 0 or c ^{x - 1} = d ^{x - 1}

But c lies in the interbal (4, 5) and d lies in the interval (2. 3)

For, c^x-1 and d ^{x - 1} to be equal, x - 1 must be 0 i.e x = 1

So, the only 2 possible real solutions are 0,1

Hence, there are only 2 possible real solutions of the given equation and they are 0 and 1 only.

Cheers!!

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