}\text{ } \\
&\text{ \ \ }&\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ }_{m-k+1}\Phi _{m-1}\left[
\left. 
\begin{array}{c}
a_{1},a_{2},...,a_{m-k},q^{\alpha } \\ 
b_{1},b_{2},...,b_{m-1}\end{array}\right\vert q,\frac{au}{vq^{\alpha }}\right] .\end{aligned}$$Hence the theorem is completely proved.

<span style="font-variant:small-caps;">Corrollary 12.</span> *Let* $a$* be a real number and* $f\left( x\right) $* be defined in terms of the* $q$*-hypergeometric function*$$f\left( x\right) =_{m-k+1}\Phi _{m-1}\left[ \left. 
\begin{array}{c}
a_{1},a_{2},...,a_{m-k}, \\ 
b_{1},b_{2},...,b_{m-1}\end{array}\right\vert q;ax\right] .$$*Then, the following identities hold.*

$\left( \mathtt{i}\right) _{q}L\left( x^{\alpha -1}f\left(
x\right) \right) \left( v\right) =\dfrac{\left( 1-q\right) ^{\alpha
-1}\Gamma _{q}\left( \alpha \right) }{K\left( \frac{1}{v};\alpha \right) }\
_{m-k+1}\Phi _{m-1}\left[ \left. 
\begin{array}{l}
a_{1},a_{2},...,a_{m-k},q^{\alpha } \\ 
b_{1},b_{2},...,b_{m-1}\end{array}\right\vert q,\dfrac{a}{vq^{\alpha }}\right] .$

$\left( \mathtt{ii}\right) _{q}S\left( x^{\alpha -1}f\left(
x\right) \right) \left( u\right) =\dfrac{u^{\alpha }\left( 1-q\right)
^{\alpha -1}\Gamma _{q}\left( \alpha \right) }{K\left( u;\alpha \right) }$ $_{m-k+1}\Phi _{m-1}\left[ \left. 
\begin{array}{l}
a_{1},a_{2},...,a_{m-k},q^{\alpha } \\ 
b_{1},b_{2},...,b_{m-1}\end{array}\right\vert q,\dfrac{au}{q^{\alpha }}\right] .$

<span style="font-variant:small-caps;">Proof</span> is straightforward from Theorem 11. Details are therefore omitted.

Let $\alpha =m=k=1$ and $a>0.$ Then, Corrollary 12 gives$$_{q}N\left( E_{q}\left( ax\right) \right) \left( u;v\right) =\dfrac{u}{vK\left( \frac{u}{v},1\right) }\text{ }_{1}\Phi _{0}\left[ 
\begin{array}{c}
q \\ 
\end{array}\left\vert q,-\dfrac{au}{qv}\right. \right] .$$Hence, it follows$$\left. 
\begin{array}{l}
\left( \mathtt{i}\right) _{q}L\left( E_{q}\left( ax\right) \right) \left(
v\right) =\dfrac{1}{vK\left( \frac{1}{v},1\right) }\text{ }_{1}\Phi _{0}\left[ 
\begin{array}{c}
q \\ 
\end{array}\left\vert q,-\dfrac{a}{qv}\right. \right] . \\ 
\left( \mathtt{ii}\right) _{q}S\left( E_{q}\left( ax\right) 