This might on first glance seem like a strange question, since it can just be said that $b=e$, hence $\log_b(a)=\ln(a)$, but hear my reasoning first.

So obviously, the "change of base" formula states that $$\log_a(x)=\frac{\log_b(x)}{\log_b(a)}$$ and when $b=e$, $$\log_a(x)=\frac{\ln(x)}{\ln(a)}$$ and then to take the derivative of $y=\log_a(x)$ from here is therefore quite easy: $$\frac{dy}{dx}=\frac{\ln(a)\cdot\frac{d}{dx}(\ln(x))-\ln(x)\cdot\frac{d}{dx}(\ln(a))}{(\ln(a))^2}$$ and since $\ln(a)$ is a constant,$$\frac{dy}{dx}=\frac{\frac{\ln(a)}{x}}{(\ln(a))^2}=\frac{\ln(a)}{x}\cdot\frac{1}{(\ln(a))^2}=\frac{1}{x\ln(a)}$$

now whilst differentiating, I can understand why the natural log is desirable, seeing as at this point, the derivative of $\ln(x)$ is already known, my question however, is: is it necessary to let $b=e$ to easily compute the derivative? The thought is very loose, so I apologise if the question is ill-worded. Any responses are appreciated.

  • $\begingroup$ The short answer is yes. $\endgroup$ – Abu Bakr Sep 13 '17 at 15:37
  • $\begingroup$ For the reason you provide, yes, and for the same reason we use$$a^x=e^{x\ln(a)}$$and appropriately apply chain rule. $\endgroup$ – Simply Beautiful Art Sep 13 '17 at 15:40

The practice of converting $\log_a(\cdot)$ to $\ln(\cdot)$ is an example of "reducing a problem to the previous case", which is usually a good idea -- there are fewer formulas to remember. (And in the case of $\log_a(x)=\frac{\ln(x)}{\ln(a)}$, you don't need the quotient rule to compute the derivative; just pull out the constant $\frac1{\ln(a)}$.)

If you're not happy about the presence of $\ln(a)$ in the derivative of $\log_a(x)$, you can use the change of base formula to express $\frac1{\ln a}$ in terms of $\log_a$: $$ \frac1{\ln a}=\frac{\log_ee}{\log_ea}=\log_a(e).$$ So an alternative way to write the derivative of $\log_a(x)$ is $$ \frac d{dx}\log_a(x)=\frac{\log_a (e)}x, $$ which uses $\log_a$ instead of $\ln$. This is correct, but a slightly less attractive form (IMO).


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