Finding a base of an exponent in mod field I have the following equation:

$$base^{exp} = number \pmod n : base,exp \in \mathbb{N}$$ and another known fact is:

$$gcd(base, number) = 1$$

$$n = 2^{2048}$$ base=13489305024865487703110255658234329747698118206959778644688156332043783846078839120693894255527894489531905012244713117142764166452312133019772171674466933769775907460046497284522592167536594047800489828714315435570429416637425443402332599055774982796405757075108551322778712959943658831605397635195107786224617525627358659165255604556424206194207190437525742567525338826878962081515333896433312311548844614323540250054093970082337500580573165008440265840792908334486258260848163001490152587781983042546491301026074736907693887630347258892882871059741621049169714319440564952700454580681894452760215968494428411686329

number=6870295205307030503255600311283969014496436297715066273709495591567561187646528069669895230912327862244474990612611625088862250315706633708998214109152824455738719595737772769297517386968692628228327225922261219083693899105983726637012353264168761696183327692619506267951701511870035935612090359086376808592001973358166067468618577312983514388332736591060901174314042634365304017788649960016991442596975922402288221898367955532116456798868804571091463566329706023967280838744359633963847966790121312196824818606244189274966061324393424041211903396020720341163472399763951106703068172772579049891895580785347369093113

I know the base and the number, how do I go about computing the exponent? 
This is related to cryptography because I'm trying to decode a message.
Solving discrete logs modulo $2^n$ is easy.

You rely on this:

If we have (for odd $e$) $$(n 2^{k+1} + 2^k + 1)^e = b \pmod{2^\lambda}$$

then either:

$$(n 2^{k+1} + 2^k + 1)^e = b \pmod{2^{\lambda+1}}$$

or

$$(n 2^{k+1} + 2^k + 1)^{e + 2^{\lambda-k}} = b \pmod{2^{\lambda+1}}$$

That can be shown because $(n 2^{k+1} + 2^k + 1)^{2^{\lambda-k}} \equiv 1 + 2^\lambda \pmod {2^{\lambda+1}}$ (binomial theorem), and so $(n 2^{k+1} + 2^k + 1)^{e + 2^{\lambda-k}} \equiv (n 2^{k+1} + 2^k + 1)^e \cdot (1 + 2^\lambda) \equiv $ $(n 2^{k+1} + 2^k + 1)^e + 2^\lambda \pmod{ 2^{\lambda+1} }$

So, you can find your exponent by starting at the bottom, $9^1 = 9 \bmod 16$, and then extending it for increasing values of $\lambda$, until you hit $\lambda = 2048$