A Simple, Yet Difficult, Math Problem

[Sebastian]

I recently came across a math problem I cannot solve, though it looks solvable. This is it:

3^x + 4^x = 5^x

As you can see, you instantly know the answer is 2(Pythagoras theorem), but how do you go about solving it?

[InfiniteNow]

What do you get when you substitute numbers for "x?" Do any of them give a solution that work?

If x=0:
3^0 + 4^0 = 5^0
1 + 1 = 1
NOPE.

If x=1:
3^1 + 4^1 = 5^1
3 + 4 = 5
NOPE.

If x=2:
3^2 + 4^2 = 5^2
9 + 16 = 25
DING!!

[C1ay]

According to Fermat's Last Theorem, 2 is the highest integer power for any such equation....

[ughaibu]

Integer values for x weren't specified in post 1.

[C1ay]

I know that. I was simply pointing out a related topic to the problem...

[Pyrotex]

Originally Posted by Sebastian

I recently came across a math problem I cannot solve, though it looks solvable. This is it: 3^x + 4^x = 5^x
...but how do you go about solving it?

You could split it into two independent (simultaneous) equations:
y = 3^x + 4^x
y = 5^x

Then, lessee here, take the Log of each equation:
R = Log y = x / Log 3 + x / Log 4
R = x / Log 5

...I think.

[Qfwfq]

And it took a while to prove Fermat's too.

[Sebastian]

Originally Posted by InfiniteNow

What do you get when you substitute numbers for "x?" Do any of them give a solution that work?

If x=0:
3^0 + 4^0 = 5^0
1 + 1 = 1
NOPE.

If x=1:
3^1 + 4^1 = 5^1
3 + 4 = 5
NOPE.

If x=2:
3^2 + 4^2 = 5^2
9 + 16 = 25
DING!!

Can you do it by any other means other than trial and error? I mean what if the problem was actually this:

4^x + 5^x = 6^x

[Qfwfq]

I'd say 99.9% it has to be done iteratively.

[max4236]

4^x+5^x=6^x, x=2.487939

I'll get back to ya later, they're calling me to dinner.