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Mathematics
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Is there any way, without guessing or read of a graph, to solve for x in an equation like 3^(x)+4^(x)=25
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>>7780411
Well in that case you have a pythagorean triple so the answer is trivial.
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>>7780414
Hurr durr x=2
I wanna know HOW to solve for x
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Hey

I tried doing 3^(x) + 3^(log3(4)*x) =25
then substituting y=3^x
Y+Y^(log3(4)) - 25 = 0
In that case you would have a "polynomial of n degree with n being non-integer, if you look up online there are some methods that allow you to calculate the number of roots and maybe a method to calculate them that I don't know of

if looking for a approximate solution you could just apply a root-finding algorithm for the last equation with two reasonable approximations of Ya and Yb, values in which the root will be encountered
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>>7780481
my guess is that if there is any, the last fermat theorem which is a^n + b^n = c^n wouldn't be a long-standing problem. Exponentials are analytic though. Meaning you can expand it into infinite sum of polynomial by Taylor series.
I also suspect if there is complex solution since the exponent of a complex number can be real.
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>>7780845
Yes, but In fermats theorem n has to be a whole number.
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>>7780845
But if you expand the exponential function into an infinite sum of polynomials using the Taylor Series how would you get the exact solution? You would get a polynomial of degree n, but the exact solution would only be found when n->inf any other n for as big as it could be would only be an approximation to the original function, right?

In that case we could just apply a root-finding algorithm, for an also approximate solution
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>>7780481
There's no step by step recipe if that's what you're asking for. It's a case by case kind of thing.
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>>7780411
let lg be log base x

lg(3^(x)+4^(x)) = lg(25) so

lg(3^x) + lg(3^x/4^x) = lg25

3 + lg((3/4)^x) = lg25

3 + 3/4 = lg(25)

x^(3+3/4) = 25 so x is the (3+3/4)th root of 25

I may have made a mistake. I know nothing about logratihms, it just looked like a logarithm problem and I googled the logarithm identities.
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>>7780411
I don't think this is analytically solvable (except in the case where n=2), further I think equations of this type can only be solved exactly if [eqn] a ~~ \text{and} ~~ b [/eqn] are multiples of each other, so [eqn] 3^x + 4^x = 23 [/eqn] isn't exactly soluble, but [eqn] 8^x + 4^x = 23 [/eqn] is.
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>>7780935
>I may have made a mistake.

You have on the second line, logs aren't linear so $\ln( 3^{x} + 4^{x}) \neq \ln( 3^{x} ) + \ln( 4^{x} )$
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>>7780948
Yea, because when they are multiples the exponent is gonna lead to a "polynomial" of n-degree with n integer and those are easier to solve, at least for a n<=4. From n=5 the solutions are harder to find analitically and in some cases I think it's impossible.. iterative methods are an option tho
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>>7780953
wikipedia says that ln(a + c) = ln(a) + ln(a/c)

However, I think I made a mistake below anyways, when I cancel the logs. That is not how that identity works.
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>>7780976
>ln(a + c) = ln(a) + ln(a/c)
I think you mean: $\ln( a+c ) = \ln (a) +\ln \left( 1+ \frac {c} {a} \right)$ what you wrote doesn't make sense since if a=c=1 then:
>ln(2)=ln(1)+ln(1)=0

Which is wrong.
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>>7780411

You didn't exclude rote memorization. In certain limited cases, it is useful to know certain things cold, exactly such as your pythagorean triple.

Other examples in pure math where rote memorization are helpful (though understanding of motivation is also needed if you're going to teach the concept or re-learn something from scratch if you need it) are the quadratic formula, exactly when matrix multiplication is defined, the formulas for manipulation of fractions, and certain simple calculus expressions involving elementary functions. These are simply learning the "words" and basic syntax, so to speak.
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There is no general solution to equations of that type in terms of elementary functions, and I can prove it.
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>>7781798
Go on then, prove it.
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>>7781763
>exactly when matrix multiplication is defined
This doesn't need to be rotely memorized. It's literally just the statement that you can only compose functions whose domain and codomain agree.
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https://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem
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>>7781836

Yes yes fine, but previous threads placed on /sci/ by teenagers have suggested otherwise.

I was simply referring to the elementary situation of an m x n matrix to the left of an n x p matrix (and let's make it even simpler: the entries are some subset of the Complexes), where their matrix product is to be evaluated (and happens to be defined, since n = n) as an m x p matrix. I like to "cancel" the two n's when I have a situation like this, as a mnemonic.

Although I don't know OTTOMH how your latter sentence pertains to the above elementary situation, when all that is implied throughout are multiplications and additions (dot products), and not composition, or the entries happen to be other objects. Unless I've missed something.
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>>7781806
It is known that the solution to $x^5+x^4=1$ has no solution in terms of elementary functions, by Galios group nonsense and the transcendence of $e$. But this equation can be written as $(e^5)^{\ln(x)} + (e^4)^{\ln(x)} = 1$, so a solution to $\ln(x)$ in terms of elementary functions would solve the original quintic in those terms.

Details left as exercise. You probably have to consider the original Galios group over the rationals adjoin e, but that won't affect the Galios group by transcendence.
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>>7781942
Every matrix comes from a linear map on a finite dimensional vector space, and vice versa. So an mxn matrix is the same thing as a linear map from k^n to k^m, with matrix multiplication corresponding to function composition.

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>>7781836
Yeah I found this suspicious to hear as well... It is like trying to rote memorize that if I have 8 hot dogs and 10 buns that I will have 2 empty buns... There's nothing unintuitive about it.
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>>7782035

>fetches my old copy of Lay, flicks through wiki
>composition... composition...

>the gist of your thing, I may abuse words slightly, is that each matrix is a map from vector space V -> W, so when you multiply them to get a third matrix it's like com-po..

OOOOOHHH, okay. I've seen rotation matrices etc but I don't recall that it was ever put in your terms of composition. If it was presented obviously I didn't remember it.
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>>7782177
Yeah. Axler's book has a section proving things about matrices from the viewpoint of linear maps that I think you might get something out of. His whole shtick is that matrices and determinants are messy and we should view things from the perspective of linear maps.