Logarithms
- Tags
- math
If some number \( a \) to the power \( x \) is \( y \), \( a^x = y \), then log to the base \( a \) of \( y \) is \( x \): \( \log_a y = x \), and equivalently \[ e^{x \ln a} = a^x \]
This can be proved through the opposite approach. Consider an equation \( \log_a x = y \), we can raise both the left and right hand sides of this by \( a \) to get \( a^{\log_a x} = a^y \). \( \log_a x \) is a number which when \( a \) is raised to that power leads to \( x \), meaning this equation is equivalent to \( x = a^y \).
Laws of Logarithms
| Name | LHS | RHS |
|---|---|---|
| Summation is Multiplication | \( \ln x + \ln y \) | \( \ln (xy) \) |
| Subtraction is Division | \( \ln x - \ln y \) | \( \ln (\frac{x}{y}) \) |
| Multiplication is Raising | \( n \ln x \) | \( \ln (x^n) \) |
| Multiplication by -1 is Division | \( - \ln x \) | \( \ln (x^{-1}) = \ln (\frac{1}{x}) \) |
And also the base conversion rule:
\begin{align} \log_a x = \frac{\log_b x}{\log_b a} \end{align}
Natural Logarithm
Is a special logarithm which is to the base of eulers number \( e \) \[ \log_e x = \ln x \]
If:
- \( y = e^x, \frac{\,dy}{\,dx} = e^x \)
- \( y = a^x, \frac{\,dy}{\,dx} = \ln (a) \times a^x \)