Count of "a"s and "b"s must be equal. Did you get it computer?

Python 3, 32 bytes

eval(input().translate(")("*50))

Outputs via exit code: Error for false, no error for True.

The string is evaluated as Python code, replacing parens ( for a and ) for b. Only expressions of the form a^n b^n become well-formed expressions of parentheses like ((())), evaluating to the tuple ().

Any mismatched parens give an error, as will multiple groups like (()()), since there's no separator. The empty string also fails (it would succeed on exec).

The conversion ( -> a, ) -> b is done using str.translate, which replaces characters as indicated by a string that serves as a conversion table. Given the 100-length string ")("*50, the tables maps the first 100 ASCII values as

... Z[\]^_`abc
... )()()()()(

which takes ( -> a, ) -> b. In Python 2, conversions for all 256 ASCII values must be provided, requiring "ab"*128, one byte longer; thanks to isaacg for pointing this out.

MATL, 5 4 bytes

tSP-

Prints a non-empty array of 1s if the string belongs to L, and an empty array or an array with 0s (both falsy) otherwise.

Thanks to @LuisMendo for golfing off 1 byte!

Try it online!

How it works

t      Push a copy of the implicitly read input.
 S     Sort the copy.
  P    Reverse the sorted copy.
   -   Take the difference of the code point of the corresponding characters
       of the sorted string and the original.

Retina, 12 bytes

Credits to FryAmTheEggman who found this solution independently.

+`a;?b
;
^;$

Prints 1 for valid input and 0 otherwise.

Try it online! (The first line enables a linefeed-separated test suite.)

Explanation

Balancing groups require expensive syntax, so instead I'm trying to reduce a valid input to a simple form.

Stage 1

+`a;?b
;

The + tells Retina to repeat this stage in a loop until the output stops changing. It matches either ab or a;b and replaces it with ;. Let's consider a few cases:

• If the as and the bs in the string aren't balanced in the same way that ( and ) normally need to be, some a or b will remain in the string, since ba, or b;a can't be resolved and a single a or b on its own can't either. To get rid of all the as and the bs there has to be one corresponding b to the right of each a.
• If the a and the b aren't all nested (e.g. if we have something like abab or aabaabbb) then we'll end up with multiple ; (and potentially some as and bs) because the first iteration will find multiple abs to insert them and further iterations will preserve the number of ; in the string.

Hence, if and only if the input is of the form anbn, we'll end up with a single ; in the string.

Stage 2:

^;$

Check whether the resulting string contains nothing but a single semicolon. (When I say "check" I actually mean, "count the number of matches of the given regex, but since that regex can match at most once due to the anchors, this gives either 0 or 1.)