Math error!

For any positive integer say $k^{ x }$ to have at most 10 digits ,it must satisfy the condition-

$k^{ x }\le { 10 }^{ 10 }-1\quad \quad \quad \quad \quad but\quad \quad \quad\quad { 10 }^{ 10 }-1\approx { 10 }^{ 10 }$ hence

${ k }^{ x }\le { 10 }^{ 10 }$

$x\ln { k } \le 10\ln { 10 }$

$x\le 10\frac { \ln { 10 } }{ \ln { k } }$

for $k=2$ , $n\le 33.219$

for $k=3$ , $p\le 20.959$

Hence $n=33$ and $p=20$

$n-p=13$