$f\left( x\right) =$ $_{r}\phi _{p}\left[ \left. 
\begin{array}{c}
a_{1},a_{2},...,a_{r} \\ 
b_{1},b_{2},...,b_{p}\end{array}\right\vert q,ax\right] $ *and* $\alpha >0.$* Then, we have*$$N_{q}\left( x^{\alpha -1}f\left( x\right) \right) \left( u;v\right) =\frac{\Gamma _{q}\left( \alpha \right) \left( 1-q\right) ^{\alpha -1}u^{\alpha -1}}{v^{\alpha }}\text{ }_{r+1}\phi _{p}\left[ \left. 
\begin{array}{c}
a_{1},a_{2},...,a_{r},q^{\alpha } \\ 
b_{1},b_{2},...,b_{p}\end{array}\right\vert q,a\frac{u}{v}\right] .  \tag{33}$$

<span style="font-variant:small-caps;">Proof</span> By taking into account $\left( 13\right) ,$ $f\left( x\right) $ is given the series representation$$f\left( x\right) =\sum\limits_{n\geq 0}\frac{\left(
a_{1},a_{2},...,a_{m-k};q\right) _{n}}{\left(
b_{1},b_{2},...,b_{m-1},q\right) _{n}}\frac{a^{n}}{\left( q;q\right) _{n}}x^{n}.$$On setting $A_{n}=\dfrac{\left( a_{1},a_{2},...,a_{m-k};q\right) _{n}}{\left( b_{1},b_{2},...,b_{m-1},q\right) _{n}}\dfrac{a^{n}}{\left( q;q\right)
_{n}}$ and using $\left( 21\right) $ and $\left( 29\right) ,$ we get $$\begin{aligned}
N_{q}\left( x^{\alpha -1}f\left( x\right) \right) \left( u;v\right) &=&\frac{\left( 1-q\right) ^{\alpha -1}u^{\alpha -1}}{v^{\alpha }}\sum\limits_{n\geq
0}A_{n}\frac{u^{n}}{v^{n}}\left( 1-q\right) ^{n}\Gamma _{q}\left( \alpha
+n\right) \\
&=&\frac{\Gamma _{q}\left( \alpha \right) \left( 1-q\right) ^{\alpha
-1}u^{\alpha -1}}{v^{\alpha }}\sum\limits_{n\geq 0}A_{n}\left( q^{\alpha
};q\right) _{n}\frac{u^{n}}{v^{n}} \\
&=&\frac{\Gamma _{q}\left( \alpha \right) \left( 1-q\right) ^{\alpha
-1}u^{\alpha -1}}{v^{\alpha }}\text{ }_{r+1}\phi _{\rho }\left[ \left. 
\begin{array}{l}
a_{1},a_{2},...,a_{r},q^{\alpha } \\ 
b_{1},b_{2},...,b_{p}\end{array}\right\vert q,a\frac{u}{v}\right] .\end{aligned}$$This completes the proof of the theorem.

By setting $p=0,\alpha =1$ and $r=0$ in $\left( 33\right) $, the above theorem leads to$$N_{q}\left( e_{q}\right) \left( ax\right) \left( u;v\right) =\frac{1}{v}\text{ }_{1}\phi _{0}\left[ q;-\left\vert q,a\frac{u}{v}\right. \right] =\frac{1}{v}\sum\limits_{n\geq 0}\left( \frac{au}{v}\right) ^{n}=\frac{1}{v-au},\text{ }\left\vert au\right\vert <v.  \tag{34}$$Further, by fixing $v=1$ and taking account of $\left( 20\right) ,$ $\left(
34\right) $ 