Geometric Bites

These workings derive the laws of base-10 logarithms from exponent laws by converting between exponential form and logarithmic form. Assume a > 0 , b > 0 , and n ≠ 0 . Product Rule log(ab) = log a + log b Let 10 x = a Let 10 y = b Therefore: log 10 a = x log 10 b = y 10 x 10 y = ab 10 x+y = ab Therefore: log 10 (ab) = x + y Substituting: log 10 (ab) = log 10 a + log 10 b Therefore: log(ab) = log a + log b Quotient Rule log(a / b) = log a - log b Let 10 x = a Let 10 y = b Therefore: log 10 a = x log 10 b = y 10 x / 10 y = a / b 10 x-y = a / b Therefore: log 10 (a / b) = x - y Substituting: log 10 (a / b) = log 10 a - log 10 b Therefore: log(a / b) = log a - log b Power Rule log(a n ) = n log a Let log(a n ) = x Therefore: 10 x = a n Taking the n-th root of both sides: (10 x ) 1/n = (a n ) 1/n 10 x/n = a Therefore: log 10 a = x / n n log 10 a = x Theref...

This article presents the rules of logarithms using complete, line-by-line derivations. Every identity is built directly from its exponential origin, without shortcuts, matching the structure of formal handwritten algebra. 1. Definition We begin with fundamental exponent facts: a⁰ = 1 ⇒ logₐ(1) = 0 a¹ = a ⇒ logₐ(a) = 1 Say: aᵐ = p Then, by definition: logₐ(p) = m Raise both sides of aᵐ = p to the power 1/m (with m ≠ 0 ): p^(1/m) = a Therefore: logₚ(a) = 1/m Since m = logₐ(p) , we obtain: logₐ(p) = 1 / logₚ(a) 2. Product Rule — Full Derivation Say: aᵐ = p and aⁿ = q Multiply: aᵐ · aⁿ = p · q Using index addition: a^(m+n) = p · q Taking logarithms: logₐ(p · q) = m + n Substitute: logₐ(p · q) = logₐ(p) + logₐ(q) 3. Quotient Rule — Full Derivation Say: aᵐ = p and aⁿ = q Divide: aᵐ / aⁿ = p / q Index subtraction gives: a^(m−n) = p / q Taking logarithms: logₐ(p / q) = m − n So: log...