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Algebra

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11 Jan 2009 13:30:40 IST

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Solving an equation

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Solve for $x\in \mathbb{R}$ the equation

$6^x + 1 = 8^x - 27^{x-1}$

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aqswdefr

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11 Jan 2009 15:55:53 IST

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( 2^x - (3^x + 3)/3) = 0

2^x =(3^x + 3)/3 ===> x= 1 , 2

Hari Shankar

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11 Jan 2009 18:43:40 IST

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sosgf, you are like some character out of Da Vinci Code man!

I mean, i know you have hit the nail on the head, but i cant undestand why you have to be so secretive and not post the complete answer. But i like your style

Soumik

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12 Jan 2009 14:15:44 IST

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Plz click on image for a better view.

While analysing the two functions and drawing their graphs, we get 2 values or solutions for sure x=1,x=2.

Graphically it can be proved that there are no other solutions....

$\ F_1$ denotes the graph of the function $\ f(x) = 6^x+1$ , & $\ F_2$ the graph of $\ g(x)=8^x-27^{x-1}$

Now lets calculate the maxima of g(x), differentiating it and using std.values of log2, log3, we have only 1 root of its derivative at approximately x = 2.4

Now from the point 1 to 2.4, suppose the function g(x) had been like this

(Click on image for better view).....

Then in the interval A o B, we would have had f(A)=f(B), and Rolle's theorem asserts us the presence of either max or min over that interval, but unfortunately the g^'(x)=0 only at x = 2.4

Thus proved only 2 solutions exist of the form 1,2.

Soumik

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12 Jan 2009 14:43:25 IST

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After all, Kaymant sir, don't give the soln now.

Dipanjan

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12 Jan 2009 22:24:53 IST

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Re:Solving an equation

Soumik

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12 Jan 2009 22:45:57 IST

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Well I made a horrible mistake - messed up with rolle entirely.

Kaymant sir, if we prove g''(x)>0 for 1<x<2.4, then I hope we are done.....

Anant Kumar

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13 Jan 2009 07:42:17 IST

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It was rather easy, it seems. However, you can also get the solutions without guessing. After, we get $3^{x-1}+1=2^x$ , we proceed as follows. Rewrite the obtained equation as

$3^{x-1}-2^{x-1}=2^{x-1}-1\qquad(1)$

Consider the function $f(t)=t^{x-1}$ defined for $t>0$ . By Lagrange's theorem, it follows that there exist $\alpha\in (1,\,2)$ and $\beta \in (2,\,3)$ such that $f(2)-f(1)=f^\prime(\alpha)$ and $f(3)-f(2)=f^\prime(\beta)$ . But $f^\prime(t)=(x-1)t^{x-2}$ . So from $(1)$ we get

$f^\prime(\beta)=f^\prime(\alpha)$

$\Rightarrow\ (x-1)\alpha^{x-2}=(x-1)\beta^{x-2}$

from where it follows that $x=1$ or $x=2$ since $\alpha\neq \beta$

Hari Shankar

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13 Jan 2009 11:36:08 IST

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One more method for solving this kind of equation which occurred to me in a similar problem.

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

Just as kaymant sir has done you can rearrange the terms and write

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

Method 1: Consider the function f(p) = p^x - (p-1)^x

When p>1, we see that for x = 0 and x =1, f'(p) = 0. For any other value of x, it is a strictly monotonic function in p.

But here we see that f(2004) = f(2002). That means x=0 or x =1

Method II: A little bit naive but intuitive reckoning that I used when I did not know convexity - concavity concepts. If you take an analogy with velocity and acceleration we can look at the function f(p) = p^x and we note that we have here $\frac{2004^x - 2003^x}{2004-2003} = \frac{2002^x - 2001^x}{2002-2001}$

Now, this function is monotonic, but further, the rate of change of the function (i.e. acceleration) will also be different in different intervals. But here, it is a constant. So look at the "acceleration" i.e. f"(p) = x(x-1)p^x-2

This must be zero, and that happens when x = 0 or x =1.

Now, if it does not occur to you to rearrange it the way sir has done, you can still solve the equation

2^x = 3^x-1+1. After observing that x=1 and x = 2 are solutions, you see that LHS and RHS are convex functions just remember that two convex curves can intersect at most two points and you have already found both of them

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