I just got introduced to logarithms and natural logarithms (I've been learning this Precalculus stuff by myself) and it all seems very confusing to me.

Logs are a simple way to find the power the base of an exponent is raised by to get the answer, for example, everyone knows that $2^3=8$ (right?), so the log of that would be $\log_2(8)=3$ because $2$ raised to the third power gives $8$.

And the same is to natural logs. Natural log (or $\ln$) is just a "compressed" way of saying $\log_e$. So since $e^1=e$, $\ln(e)=1$.

Question: Is there an easy way to memorize the places where you write the power, and the outcome?

Or, we have:

If $$b^a=c$$ then $$\log_b(c)=a$$

I'm just wondering if there's an easy way to memorize that, because remembering where I put the powers is the difficult part for me.

Also, just a quick little question, you don't have to answer this.

  • Why is Natural Logarithm denoted as $\ln$? The "l" is first, so shouldn't it be "Logarithm natural"?
  • 1
    $\begingroup$ I usually think that "$\log_b$ is the inverse function of $b^{\bullet}$". Or, that $\log_b(b^x)=x$. $\endgroup$ – user228113 Oct 19 '16 at 21:41
  • 1
    $\begingroup$ The big number comes after the $\log$. The number below the exponent ($b$) also goes below the $\log$. And of course, the special number, what it's raised to the power of, is the answer to the logarithmic equation. $\endgroup$ – Skeleton Bow Oct 19 '16 at 21:43
  • 1
    $\begingroup$ you can "logarithmization" both side as $b^{ a }=c\\ \log _{ b }{ { b }^{ a } } =\log _{ b }{ c } \\ a=\log _{ b }{ c } $ $\endgroup$ – haqnatural Oct 19 '16 at 21:46
  • $\begingroup$ It's easier when you remove $c$ altogether. $\log_b{b^a}=a$. $\endgroup$ – Dean C Wills Oct 19 '16 at 23:07
  • $\begingroup$ I'll answer the last question. Usually it's because of Latin: "Pietro Mengoli and Nicholas Mercator called it logarithmus naturalis a few decades before Newton and Leibniz developed calculus." $\endgroup$ – Cehhiro Oct 20 '16 at 1:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.