For pset 1, problem 2, I am a little confused as to taking the logs of prime numbers. Why is the log(3) equal to 1.0986...?

Question

e.mccormick · Answer

$y=b^x$ solve for x.  OK, so we take $\log_b$ of both sides.

$\log_by=\log_bb^x$

$\log_by=x\log_bb$

$\log_by=x1$

$\log_by=x$

As you can see, logs and exponents are related.

Now, some people use $\log$ to mean the natural logarithm, $e$.  Others use it to mean base 10.  Too bad they don't all use ln to mean the natural logarithm.

So what is $e^{1.0986...}$?  Well, can't 100% accurately do that.  But if I take an approximation of e, I get:

$2.71828182846^{1.0986}= 2.9999631342233446407228273282365$

That is very close to 3 and hopefully you can see it follows the relationship shown.

anonymous · Answer

Hi e.mccormick, 

Thanks for your response to my question. 

Yes, I was confused because I assumed that the log was base 10 instead of base e. I guess I am still new to this. 

Thanks again for clarifying.