Coaching Centre
Logarithmic functions to other bases
Changing log_a x to base e and differentiating logarithmic functions to any base, including with the chain rule.
The derivative
By the change-of-base formula, $\log_a x = \dfrac{\ln x}{\ln a} = \dfrac{1}{\ln a} \times \ln x$. The factor $\dfrac{1}{\ln a}$ is just a constant.
Differentiate (a) $\log_2 x$ and (b) $\log_{10}(3x + 1)$.
Find the gradient of $y = \log_3 x$ at its $x$-intercept.
$\log_a x = $
The change-of-base formula.
$\dfrac{d}{dx}\log_2 x = $
$\dfrac{1}{\ln 2} \times \dfrac1x$.
Find the gradient of $y = \log_3 x$ at $x = 1$. Type $\ln 3$ as ln3.
$\dfrac{1}{1 \times \ln 3}$.
$\dfrac{d}{dx}\log_{10}(3x + 1) = $
$\dfrac{f'(x)}{f(x)\ln a}$.
Find $\log_2 10$ to 3 decimal places.
$\dfrac{\ln 10}{\ln 2}$.
Find the gradient of $y = \log_{10} x$ at $x = 10$.
$\dfrac{1}{10\ln 10}$.
Find the gradient of $y = \log_2 x$ at $x = 4$.
$\dfrac{1}{4\ln 2}$.
For $x > 0$, $\dfrac{d}{dx}\log_2 x^2 = $
$\log_2 x^2 = 2\log_2 x$.
Why is the derivative of $\log_e x$ simply $\dfrac1x$?
The factor $\dfrac{1}{\ln a}$ becomes $1$.
New numbers every time. Keep going until you get five in a row.