Question

Challenge: Given that \(\Large a^x + a^y = a^z\), where \(\Large a \neq 0\)
Find z in term of x and y.

Answer 1

Good question. Maybe these guys can help you? @UnkleRhaukus @.Sam. @AravindG @ganeshie8 @RadEn @goformit100 @terenzreignz @Preetha @inkyvoyd @nubeer @genius12

Answer 2

take log on both sides

Answer 3

Does this count?
\[\frac{\ln(a^x+a^y)}{lna}=z\]

Answer 4

:D ^^

Answer 5

I hope he don't mind a.

Answer 6

^I wanted to do that too, but it seemed too good to be true :(

Answer 7

x=y=z=2

Answer 8

\[2a^2=a^2 ??\]

Answer 9

1=2 ?!

Answer 10

look like someone broke math.

Answer 11

does a=a?

Answer 12

why shouldnt it?

Answer 13

i would like to withdraw my attempt

Answer 14

np :)

Answer 15

how about
a=2, x=1, y=1, z=2

z=x+y

Answer 16

Why are you going for specific cases here @UnkleRhaukus ?

Answer 17

"Find z in term of x and y". How could constants come as the answer?

Answer 18

Fermat's Last thm

Answer 19

=.=

Answer 20

and Beal's conjecture (unproved thm)

Answer 21

whats Beal's conjectur

Answer 22

http://en.wikipedia.org/wiki/Beal's_conjecture

Answer 23

So it's like Fermat's Last Theorem, except that all bases are same instead of exponents, like other way around, right?

Answer 24

yes. So, can anybody invite Andrew Weyl to Open Study to answer this?

Answer 25

umm who's Andrew Weyl?

Answer 26

http://en.wikipedia.org/wiki/Andrew_Wiles

Answer 27

Oh... haha