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Differential Calculus

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13 Mar 2008 18:29:41 IST

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Equation

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Find all solutions to 2^x+5^x = 3^x+4^x

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Arnav Guha

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13 Mar 2008 18:43:58 IST

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x=0

ha ha

Rahul Raghavendra

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13 Mar 2008 18:45:29 IST

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0 and 1

Hari Shankar

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13 Mar 2008 18:54:33 IST

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can you prove no other solutions exist?

Rahul Raghavendra

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13 Mar 2008 18:58:30 IST

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trying it

Gaurav |spideyunlimited| Ragtah

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13 Mar 2008 18:59:05 IST

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0, 1.

Ajay Antony

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13 Mar 2008 19:08:58 IST

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0,1
by dividing by 4^x,
(1/2)^x + (5/4)^x= 1+ (3/4)^x
=> (5/4)^x= 1+ (3/4)^x-(1/2)^xRHS is a decreasing fn while LHS is an inc fn therfore for except at x=0,1 there exists no soln

Gaurav |spideyunlimited| Ragtah

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13 Mar 2008 19:09:05 IST

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2^x+5^x = 3^x+4^x2^x + 5^x = 3^x + 2^2x5^x - 3^x = 2^2x - 2^x
5^x - 3^x = 2^x ( 2^x - 1)

2^x = (5^x - 3^x) / ( 2^x - 1)
Now for all powers of 5 and 3 which are not 0 and 1, the numbers will be odd...
For x not equal to 0 or 1: 5^xis odd, 3^xis odd and 2^x-1 is odd ( even number -1 )
but in LHS we have an even number.

So equation is only satisfied by x = 0, 1 which we can see.

Ajay Antony

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13 Mar 2008 19:09:42 IST

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Pls rate if im correct .

Hari Shankar

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13 Mar 2008 19:14:12 IST

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annihilator, can you elaborate on that solution. If one is strictly increasing and the other strictly decreasing, so how come 2 solns. Nice approach, but sth needs to be added.

spidey, its not just integer solutions.

anchit saini

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13 Mar 2008 19:33:47 IST

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sir isn't this the same kind of qn as this--

http://www.goiit.com/posts/list/differenciation-exponential-eqn-44485.htm

Ajay Antony

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13 Mar 2008 19:40:49 IST

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It is so because for those two values their sum can add upto 5/4 but in greater powers of x their value bcums increasingly samller.

Hari Shankar

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13 Mar 2008 22:58:02 IST

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anchit is right. This has been solved sometime earlier. Yet there is one more approach I want to show you.

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

Consider the function f(x) = x^a where a is a constant.

Now f'(x) = ax^a-1and f"(x) = a(a-1)x^a-2.

Now f"(x) is like acceleration, it measures the rate at which the function is increasing or decreasing. If a

0,1 then it is a non-zero number.

But here the rate of increase is zero. This means f"(x) = 0. This means a=0 or a=1.
finito.

@annihilator: sweetheart, it is not for me that I ask for elaborations, but for your friends who will learn something useful.

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