Determining the rate equation from a reaction where one reactant is in excess
There is no real need to invoke a "$k^{app}$". 

Let the rate be $r$, so:

$$r=k[\ce{A}][\ce{B}]^n$$

Since as $\ce{A}$ is in large excess:

$$[\ce{A}]=C(A)\gg[\ce{B}] \implies r=kC(A)[\ce{B}]^n$$

So $[\ce{A}]$ doesn't vary perceptibly during the rate measurement interval.

Plot the value of $\ln r$ versus $\ln [\ce{B}]$ in the knowledge that:

$$\ln r=\ln k + \ln C(A) + n\ln [\ce{B}]$$

So the gradient of that graph is $n$. 

And the intercept of that graph is $\ln k + \ln C(A)$. Since as $C(A)$ is known, $\ln k$, and thus $k$, is also known.