# Solve 2^x=13mod 3^4

Solve ${2}^{x}=13mod{3}^{4}$
I know $\mathrm{log}13=30mod{3}^{4}$ and $\mathrm{log}16=15mod{3}^{4}$
I've tried subbing $\mathrm{log}13/\mathrm{log}16$ for $2$ but I am not sure what to do next.
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

giosgi5
$\begin{array}{}\text{(1)}& {2}^{x}=13mod{3}^{4}\end{array}$
means that ${2}^{x}={3}^{4}k+13=81k+13$ for some $k\in \mathbb{Z}$
For $81k+13$ to equal ${2}^{x}$, for some $k,x\in \mathbb{Z}$, $k$ must be odd ($81k+13$ must be even).
For $k=1$, it exists no $x\in \mathbb{Z}$ such that $81+13={2}^{x}$
But for $k=3$, you get $81\cdot 3+13=256={2}^{8}={2}^{x}$
Hence, $x=8$ is one (integer) solution of (1).
###### Did you like this example?
hogwartsxhoe5t
In this case you're looking for ${\mathrm{log}}_{2}13\phantom{\rule{0.444em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}{3}^{4}\right)$, and you can write
${\mathrm{log}}_{2}13=\frac{\mathrm{log}13}{\mathrm{log}2}\phantom{\rule{1em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}{3}^{4}\right)$
Moreover, we know $\mathrm{log}16=\mathrm{log}{2}^{4}=4\mathrm{log}2$. So $\mathrm{log}2=\frac{1}{4}\mathrm{log}16$. Substituting this yields
${\mathrm{log}}_{2}13=\frac{\mathrm{log}13}{\frac{1}{4}\mathrm{log}16}=4\frac{\mathrm{log}13}{\mathrm{log}16}\phantom{\rule{1em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}{3}^{4}\right)$