Given a positive real number \( x \) and a base \( b > 0 \) with \( b \neq 1 \), the logarithm of \( x \) with base \( b \) ( \( \log_b(x) \) ), is the exponent \( y \) to which we must raise \( b \) to obtain \( x \). Formally:

\[ \log_b(x) = y \iff b^y = x \]

Logarithmic Properties

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: \(\log_b(x^n) = n \cdot \log_b(x)\)

Let \( y = \log_b(x) \). Since \( b^y = x \), raising both sides to the power of \( n \) and applying the property \( (b^y)^n = b^{ny} \), we get:

\[ (b^y)^n = x^n \implies b^{ny} = x^n \]

Applying the logarithmic definition 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 obtain the product rule: \[ \log_b(x \cdot y) = \log_b(x) + \log_b(y) \]

Quotient Rule: \(\log_b \left( \frac{x}{y} \right) = \log_b(x) - \log_b(y)\)

Again, let \( m = \log_b(x) \) 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: \(\log_b(x) = \frac{\log_c(x)}{\log_c(b)}\)

Taking the logarithm with base \( c \) on both sides of \( b^y = x \), we get: \[ y \cdot \log_c(b) = \log_c(x) \implies y = \frac{\log_c(x)}{\log_c(b)} \]

and thus: \[ \log_b(x) = \frac{\log_c(x)}{\log_c(b)} \]