## Content

### Derivatives of general logarithmic functions

Consider a logarithmic function $$f(x) = \log_a x$$, where $$a>1$$ is a constant. By using the change of base rule, we can write $$f(x)$$ in terms of the natural logarithm, and then differentiate it:

$f(x) = \log_a x = \dfrac{\log_e x}{\log_e a}$ and so, since $$\log_e a$$ is a constant, the derivative is $f'(x) = \dfrac{1}{x \, \log_e a}.$

#### Example

Differentiate $$f(x) = \log_2 (3-7x)$$.

#### Solution

Since

$f(x) = \dfrac{\log_e(3-7x)}{\log_e 2},$ noting that $$\log_e 2$$ is just a constant and using the chain rule, we have \begin{align*} f'(x) &= \dfrac{1}{\log_e 2} \cdot \dfrac{1}{3-7x} \cdot (-7)\\ &= \dfrac{-7}{(3-7x) \, \log_e 2}. \end{align*}

Exercise 10

Consider $$f(x) = \log_a (xy)$$ and $$g(x) = \log_a x + \log_a y$$, where $$a>1$$ and $$x,y$$ are positive. As our notation suggests, think of $$x$$ as a variable and $$y$$ as a constant.

1. Show that $$f'(x) = g'(x)$$.
2. Show that $$f(1) = g(1)$$.
3. Conclude that $$f(x) = g(x)$$ for all positive $$x$$.

This gives an alternative proof of one of the logarithm laws.

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