What is the standard error of the mean of an exponential distribution of the form $Ae^{Bx}$ with N measurements? What is the standard error of the mean of an exponential distribution of the form $Ae^{Bx}$ with $N$ measurements?
 A: I will recast your problem in slightly more formal language.  Let $X$ have density function $Ae^{Bx}$ for $x \ge 0$, and $0$ for $x<0$.  Then $B$ has to be negative, say $B=-\lambda$, where $\lambda$ is positive.  And for the integral of the density function from $0$ to infinity to be $1$, you need $A=\lambda$.
What is the mean of $X$? It is a standard fact that this is
$$\int_0^\infty x\lambda e^{-\lambda x}dx$$
Integrate (by parts), or if you are in a pre-integration phase, accept the fact that this integral is equal to $1/\lambda$.  
What is the variance of $X$?  By a standard fact, this is $E(X^2)-(E(X))^2$. For $E(X^2)$, integrate $x^2$ times your density function from $0$ to infinity.  After a while, you will find that the variance of $X$ is $1/\lambda^2$.  Or else accept from your notes that this is the case.
Now let $X_1, X_2,\dots,X_N$ be independent exponentially distributed all with the same $\lambda$.  Your sample mean is the random variable
$$Y=(X_1+X_2+\cdots +X_N)/N$$
Thus $Y$ is a constant ($1/N$) times a certain sum.
The variance of $Y$ is $(1/N^2)$ times the variance of $X_1+X_2+\cdots +X_N$. But the variance of a sum of independent random variables is the sum of the variances, so the variance of $Y$ is
$$(1/N^2)(N/\lambda^2)$$
which is $1/(N\lambda^2)$.
For the standard error, aka standard deviation, of $Y$, take the square root of the variance.
You will get
$$\frac{1}{\lambda\sqrt{N}}$$
You may have to adapt the reasoning to the particular tools, and language, that you are expected to use.  I hope that will not be difficult. 
Note: Maybe your notes/text just tell you that the standard error of the exponential is $1/\lambda$, and that the standard error of a sample mean is $1/\sqrt{N}$ times the standard error of any one of your experiments.  Then the answer $1/(\lambda\sqrt{N})$ comes with essentially no calculation needed. 
A: $Ae^{Bx}$ isn't a probability density (supported on the positive real numbers) unless it is non-negative and integrates to $1$ which only happens when $B < 0$ and $A = |B|$, which is the "classical" exponential distribution and the answer of user6312 is apt. 
