The Math Thread!

I love maths. I used to ace every test, then I stopped caring and got 90-95% on them. The people at my school are idiots though, they average stuff like 56% on a simple shapes test. SHAPES. How do you stuff that up so badly? TELL ME.
 
Hey Zeto, I'm preeeetty sure that the equation you have there has no known solution... How'd you end up getting it?

My math story: I never paid attention at secondary school, but did alright enough to proceed to take Math and Further Math at A level. I was at the "I really don't care" years of my life back then, and didn't learn much at all. I actually remember getting 0% on a math test! Once I left full time education and entered full time work, I found the space I needed to actually appreciate how nifty a lot of mathematics (and physics) was. I was far happier studying it on my own terms. I worked quiet night shifts, so I'd usually bring my own notes to work and study to pass the time. After a year of that, I was lucky enough to get into University, where I learned a lot more. Uni is now over for me, but I still study most of my free time. I just like math. I'm combing through two math-heavy books atm

I guess the moral of that story is to not be discouraged if you do badly in math at school. School is usually oppresive in the sense that you spend so much time trying to ingest the sylabus that you forget that it can actually be fun. There is little room for creativity, little room for freedom, when it comes to mainstream education. Do your own thing, study weird and obscure things, piss about with equations and see how far you can take them. Try and break them, even.
 
Zeto said:
Hey, i have a particular math problem here. During our study of exponential and logarithmic functions, I got a special equation i cannot solve. It s stated in the simple form as below:

a^x=x
or
a^x+x=0
a being obviously a constant.

I wonder if it s possible to solve this equation.
Click to expand...

Are these two formulas supposed to be derived from each other? If so, I think you meant;

a^x = x
and;
a^x - x = 0

The only circumstance I see this possible is; if a = x

Even if if you knew the x, the a would most likely be a irrational number, and changing depending on your x. Since a is a constant, I cannot see a possibility of solving it.
 
^yes, my bad, typo.
Actually, these two functions do not touch (f(x)=a^x and g(x)=x; I proved that) but if g(x)=x+b, then, there s a possibility.
But a way to find the solution by using algebraic meanings has yet come to my mind...
 
Zeto said:
^yes, my bad, typo.
Actually, these two functions do not touch (f(x)=a^x and g(x)=x; I proved that) but if g(x)=x+b, then, there s a possibility.
But a way to find the solution by using algebraic meanings has yet come to my mind...
Click to expand...

Well, the point is, that if a is a constant (in a^x = x), it is impossible to solve it, unless you actually know the constant. Since the value of x depends on a. As the value of a changes, so does the x.
 
^Really? if you know the constant a, would you be able to solve f(x)=g(x)?

Ok, what if a=2 and b=1?
I know the answer (it s 0), but how to solve it algebraically is another matter.
 
Zeto said:
^Really? if you know the constant a, would you be able to solve f(x)=g(x)?

Ok, what if a=2 and b=1?
I know the answer (it s 0), but how to solve it algebraically is another matter.
Click to expand...

Well, if your assumption is f(x) = g(x) then;

a^x = x + b

2^x = x + 1

2log(x+1) = x

x = 0 V x = 1

I don't think this fully satisfies your answer, I'll come back on it when I have some more time.
 
^meh... You know, what if a and b are huge or minute numbers?
If I ask my question in another way, can x+logx=something be simplified into x=something else, in order to find the unown value?

I have to admit I have browsed through a lot of sites, forums, but still no answers. I can t believe there has not yet been a problem where one has to find the solution of an exponential and a linear function...
 
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