more than one way ...

RauLiTo · 2006-11-27 06:52:25

solve this :
3 ^ n = 2 ^ ( n + 1 ) + 1

one of the solution is n = 2 ... but i want the way it can solved and if its possible more than one way

thanks lovely members

luca-deltodesco · 2006-11-27 07:32:34

i dont know how you would solve this mathematically, you cannot use logarithms because of the nature of the right hand side here.

RauLiTo · 2006-11-27 07:36:48

thats the problem ... but nothing is impossible in math !

luca-deltodesco · 2006-11-27 07:39:13

apparently, you can use calculus (reference to toast) but im not sure how you would do that

RauLiTo · 2006-11-27 07:41:02

ooh okay thanks anyways ... guys anybody can help please ?

RauLiTo · 2006-11-27 14:49:03

nobody ?!

Zhylliolom · 2006-11-27 19:40:01

The calculus method should provide some insight into the problem. Define the function f(x) := 3[sup]x[/sup] - 2[sup]x+1[/sup] - 1. The zeros of this function will correspond to solutions to your problem. By your inspection already, we know that x = 2 is a zero. Now our goal is to see if any other zeros exist. First, find the extrema of the function. Taking the derivative, we have f'(x) = 3[sup]x[/sup]ln 3 - 2[sup]x[/sup]2 ln 2. Setting the derivative equal to zero, we find that at the extreme values 3[sup]x[/sup]ln 3 = 2[sup]x[/sup]2 ln 2. Take the natural logarithm of both sides and solve for x to get x = (ln(2 ln 2) - ln(ln 3))/(ln 3 - ln 2) ≈ 0.5736. So there is only one extreme value for this function, and it is a minimum (I leave it to you to test that it is indeed a minimum). Now investigate the end behavior of the function: will it ever cross the x-axis again? There are no other maximums or minimums, so we know that f(x) can't jump above the x-axis and then under it (or vice-versa). So let us look at what happens to the function as x approaches -∞ and also when x approaches +∞.

What does this tell us? We know that f(x) is negative around values such that x < 2. The minimum of f, ~f(0.5736) was also a negative value. Then as x gets increasingly smaller f(x) approaches -1; this tells us that f(x) will never cross the x-axis as x -> -∞. Why is this? If f(x) went above the x-axis for some very negative value of x, then it would need to once again go below it so that as x -> -∞ f(x) -> -1. But this would require f(x) to have a maximum somewhere, which it does not. Likewise, we know that f(x) is positive around values such that x > 2. The limit tells us that f(x) will just get bigger and bigger in that direction. So for a zero to exist past 2, f(x) would need to dive below the x-axis and then rise back up towards ∞ in the limit, which would create a maximum and a minimum. But once again, we know that f(x) only has one minimum and this minimum has already been determined. From this evidence, we may conclude that f(x) has only one zero, at x = 2, and thus the equation 3[sup]n[/sup] = 2[sup]n+1[/sup] + 1 only has the solution n = 2.

luca-deltodesco · 2006-11-27 20:37:05

that makes sense, but how would you go about deriving the root 2 from the original equation since you cant use logarithms to derive that root? How would do that using the 'calculus way'

Zhylliolom · 2006-11-27 20:52:20

For an arbitrarily complicated function (so that we may not be able to analytically find exact roots), a 'calculus way' to find the (approximate, in most cases) roots of the function is Newton's method. From an initial guess at the root x[sub]1[/sub], we may take successive approximations of the form x[sub]n+1[/sub] = x[sub]n[/sub] - f(x[sub]n[/sub])/f'(x[sub]n[/sub]). If the root is r, then

Last edited by Zhylliolom (2006-11-27 20:54:36)

luca-deltodesco · 2006-11-27 21:54:15

ah, i see.

so in this case

take a guess at 4

is this method better than the +/- one, where you take two points on opposing sides of the x axis, and iteratively subdivide to gain another 1dp of accuracy? This method does seem to suffer from the same fatal flaws like the iteration one where you can only really find one root, and doesnt always work?

Last edited by luca-deltodesco (2006-11-27 21:59:23)

RauLiTo · 2006-11-28 06:00:12

although i dont know what are you talking about maybe because i didnt study that yet thank you all guys u really care ...

who can solve this in easy way :

2 ^ x = x ^ 2

luca-deltodesco · 2006-11-28 08:09:28

that again is like the first, it cant be solved analytically

finding it via subdivision of the graph algorithmically, i get roots around about (when searching from -10 to 10 only)

-0.7666473388671875
1.9999847412109375
3.9999237060546875

im guessing second two would diverge to be 2 and 4, not sure about that first one though, although graphing the function, it seems correct.

Last edited by luca-deltodesco (2006-11-28 08:26:07)

Toast · 2006-11-28 08:11:59

luca-deltodesco wrote:

ah, i see.
so in this case

take a guess at 4
is this method better than the +/- one, where you take two points on opposing sides of the x axis, and iteratively subdivide to gain another 1dp of accuracy? This method does seem to suffer from the same fatal flaws like the iteration one where you can only really find one root, and doesnt always work?

ye............. I would never have been able to explain this to you, good thing you figured that out yourself, I really don't understand any of this.

Last edited by Toast (2006-11-28 14:16:19)

Naresh · 2006-12-02 04:19:35

the best way is using graphs.Use incresing part to find the extremums and plot the graph.You will get an extremely accurate answer.

Math Is Fun Forum

#1 2006-11-27 06:52:25

more than one way ...

#2 2006-11-27 07:32:34

Re: more than one way ...

#3 2006-11-27 07:36:48

Re: more than one way ...

#4 2006-11-27 07:39:13

Re: more than one way ...

#5 2006-11-27 07:41:02

Re: more than one way ...

#6 2006-11-27 14:49:03

Re: more than one way ...

#7 2006-11-27 19:40:01

Re: more than one way ...

#8 2006-11-27 20:37:05

Re: more than one way ...

#9 2006-11-27 20:52:20

Re: more than one way ...

#10 2006-11-27 21:54:15

Re: more than one way ...

#11 2006-11-28 06:00:12

Re: more than one way ...

#12 2006-11-28 08:09:28

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#13 2006-11-28 08:11:59

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#14 2006-12-02 04:19:35

Re: more than one way ...

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