Given a positive real number \( x \) and a base \( b > 0 \) with \( b \neq 1 \), the logarithm of \( x \) to the base \( b \) ( \( \log_b(x) \) ) is the exponent \( y \) to which \( b \) must be raised to obtain \( x \). Formally:
\[ \log_b(x) = y \iff b^y = x \]
Identity
- \( b^{\log_b(a)} = a \)
If \( x = \log_b(a) \), then by definition \( b^x = a \). Substituting \( x \) with \( \log_b(a) \), we get the desired equality: \[ b^{\log_b(a)} = a \]
Exponent Rule
- \(\log_b(x^n) = n \cdot \log_b(x)\)
Consider \( y = \log_b(x) \). From \( b^y = x \), we raise both sides to the power of \( n \) and apply the property \( (b^y)^n = b^{ny} \):
\[ (b^y)^n = x^n \implies b^{ny} = x^n \]
Applying the definition of the logarithm to \( x^n \), we obtain:
\[ \log_b(x^n) = n \cdot \log_b(x) \]
Product Rule
- \(\log_b(x \cdot y) = \log_b(x) + \log_b(y)\)
Let \( m = \log_b(x) \) and \( n = \log_b(y) \). By definition, we have \( b^m = x \) and \( b^n = y \). Since \( \log_b(b^{m+n}) = m + n \), we get the product rule: \[ \log_b(x \cdot y) = \log_b(x) + \log_b(y) \]
Quotient Rule
- \(\log_b \left( \frac{x}{y} \right) = \log_b(x) - \log_b(y)\)
Let again \( m = \log_b(x) \) and \( n = \log_b(y) \). Since \( \frac{x}{y} = b^m \cdot b^{-n} = b^{m-n} \), the logarithm of the quotient becomes: \[ \log_b \left( \frac{x}{y} \right) = \log_b(b^{m-n}) = m - n \implies \log_b \left( \frac{x}{y} \right) = \log_b(x) - \log_b(y) \]
Change of Base
- \(\log_b(x) = \frac{\log_c(x)}{\log_c(b)}\)
If we take the logarithm to base \( c \) of both sides of \( b^y = x \), we obtain: \[ y \cdot \log_c(b) = \log_c(x) \implies y = \frac{\log_c(x)}{\log_c(b)} \]
Therefore: \[ \log_b(x) = \frac{\log_c(x)}{\log_c(b)} \]
