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...

Two of the most important results in differential calculus are d/dx(ln x) = 1/x d/dx(aˣ) = aˣ ln a. These formulas are closely connected. One describes the derivative of the natural logarithm, while the other gives the derivative of an exponential function with positive base. Together, they reveal the deep relationship between logarithms, exponentials, inverse functions, and differentiation. 1. Derivative of the natural logarithm Begin with ln x = y. This is equivalent to eʸ = x. Differentiate both sides with respect to y: dx/dy = eʸ. Now invert this result: dy/dx = 1/eʸ. Since eʸ = x, substitute back: dy/dx = 1/x. Therefore, d/dx(ln x) = 1/x, for x > 0. So the derivative of the natural logarithm is the reciprocal of x. 2. Derivative of the exponential function aˣ Now let y = aˣ, where a > 0 and a ≠ 1. Take logarithms to base a: logₐ y = x. Now apply the change-of-...

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...