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Algebra

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17 Aug 2007 19:17:16 IST

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R U A GENIUS, PROVE IT !!!

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4 ^x + 3 ^x = 5 ^x + 2 ^x has only 2 real solutions.

If anyone solves this, to me he / she will be a genius.

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Nishant Dey

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17 Aug 2007 19:33:16 IST

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First by looking at the equations, we see that x=0 and x=1 are two real solutions

Now 3^x-2^x=5^x-4^x
Expand with binomial theorem and get the following
xC1(4-2)+xC2(4^2-2^2)+...........+xCx(4^(x-1)-2^(x-1))=0
This has no real solutions
So x=1 and x=0 are the only solutions

Akansha

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17 Aug 2007 19:34:31 IST

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x=0,1(can b done by simply seeing the eqn)
f(x)=4 ^x + 3 ^x - 5 ^x - 2 ^x
f(1)=0
divide the whole eqn by 5^x
lim as x-->inf f(x)= -1
f(x) varies frm +-ve to -ve after x=0 so it cuts the x axis at only 1 pt
ie only two real solns

Titun

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Joined: 23 Dec 2006

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17 Aug 2007 19:44:39 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!!

Rajat Sen

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17 Aug 2007 21:31:15 IST

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the given equation can be written 2^2x - 2^x +3^x-5^x=0
this is a quadratic in 2^x
the roots are (1+/-(root(1+4(5^x-3^x)))/2
x will be real only if this is rational .
so x can be 1 and 0 and 2^x can be 2and 1.
these are the two solutions because a quadratic cannot have more than two roots.

Avirup Dasgupta

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18 Aug 2007 07:44:44 IST

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THe given equation can be written as (2^x)²-2x+3^x-5^x=0

This is a quadratic

calculate the roots.

For x to be real, x can be either equal to 1 or 0

Hence proved.

Note that this equation, being a quad. xannot have more than two solutions of 2^x and hence two solutions of x.

please rate me..............

Rajat Sen

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18 Aug 2007 13:56:39 IST

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Why are you copying my solution metal.

Rajat Sen

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19 Aug 2007 12:01:44 IST

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Why doont i get rated if the answer is correct?

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