Human Calculator Kevin Sun

Level 2

Find $log_{37}177917621779460413$

1 solution

Saurabh Mallik
May 11, 2014

Firstly, we need to find the prime factors of $177917621779460413$ .

$177917621779460413 = 37^{11}$

So, $log_{37}177917621779460413$ can be written as: $37^{x}=177917621779460413$

$37^{x}=37^{11}$

$x=11$

Thus, the answer is: $log_{37}177917621779460413 = \boxed{11}$

How would you factor that huge number as 37^11???

Nathan Singiri - 6 years, 3 months ago

Ask wolfram alpha ;-).

Maurice van Peursem - 2 years, 9 months ago