You haven't logged in in years, but seeing as someone just edited the tags on this question maybe other people will find it.
I think "it's not obvious" is being a little too pessimistic. Insofar as exponentiation is repeated multiplication, and multiplication in a tropical semiring is addition, it makes sense to define the tropical exponential "$x^n$" to be the linear function $nx$, at least for integer $n$. Then you extend to rational exponents $\frac{p}{q}$ by taking unique roots; that is, you ask that the "tropical exponential" satisfy the property that "$\left( x^{\frac{p}{q}} \right)^q$" is equal to "$x^p$" which forces the exponential to be $\frac{p}{q} x$. And then, just as for the ordinary exponential, the result for real exponents is determined by continuity.
The main conceptually tricky thing about this argument, to my mind, is that you are forced to make clear distinctions between the base and the exponent: while the base is an element of the tropical semiring the exponent is an ordinary integer or rational number or real number, regarded as a ring with the usual ring operations. This interpretation is necessary to make the exponent rules continue to be true: for example, we have "$x^{n+m} = x^n x^m$" but while the addition on the LHS is ordinary the multiplication on the RHS is tropical, since this identity really says
$$(n+m)x = nx + mx$$
and similarly we have "$(x^n)^m = x^{nm}$" which says
$$m(nx) = mnx.$$
A closely related argument that comes to the same conclusion is the following. You can pass from ordinary addition and multiplication to tropical addition and multiplication as follows: for a real parameter $t$ consider the family of operations
$$x +_t y = \frac{\log (e^{tx} + e^{ty})}{t}$$
$$x \times_t y = \frac{\log (e^{tx} e^{ty})}{t}$$
obtained by conjugating ordinary addition and multiplication by the map $x \mapsto e^{tx}$. The second operation is just always equal to tropical multiplication $x + y$. The first operation approaches tropical addition $\text{min}$ as $t \to - \infty$. Intuitively speaking this says that the tropical operations describe the asymptotic behavior of ordinary addition and multiplication applied to exponentials with very large (negative in this case, but you can use a different tropical semiring with $\text{max}$ corresponding to the positive case) exponents.
So tropical exponentiation ought to be given by conjugating ordinary exponentiation $x^r$ by the same map. This gives
$$\frac{\log e^{txr}}{t} = rx$$
as expected (and as with multiplication we don't even need to take the $t \to - \infty$ limit). And again we see the key feature that the base $x$ is treated as an element of the tropical semiring but the exponent $r$ is not.
