# Just the maths - teaching slides by Hobson A.J.

By Hobson A.J.

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2 ILLUSTRATIONS 1. logB 1 = 0 simply because B 0 = 1. 2. logB B = 1 simply because B 1 = B. 3. logB 0 doesn’t really exist because no power of B could ever be equal to zero. 3 USEFUL RESULTS (a) For any positive number x, x = B logB x. Proof In x = B y , replace y by logB x. (b) For any number y, y = logB B y . In y = logB x, replace x by B y . 4 PROPERTIES OF LOGARITHMS (a) The Logarithm of Product. q = logB p + logB q. B logB q = B logB p+logB q . (b) The Logarithm of a Quotient p logB = logB p − logB q.

5 1 ≡ , 25 + 15x 5 + 3x assuming that x = − 53 . 2. 4x 4 ≡ , 2 3x + x 3x + 1 assuming that x = 0 or − 13 . 3. x+2 x+2 1 ≡ ≡ , x2 + 3x + 2 (x + 2)(x + 1) x + 1 assuming that x = −1 or − 2.

3 × 34 = 102 and 2 × 34 = 68. 170 = 102 + 68 2. 25. 25 = 125. 75 + 125. 1 INDICES (a) Positive Integer Indices Let a and b be arbitrary numbers Let m and n be natural numbers Law No. 1 am × an = am+n Law No. 2 am ÷ an = am−n assuming m greater than n. Note: am am = 1 and am am = am−m = a0. Hence, we define a0 to be equal to 1. Law No. 3 (am)n = amn ambm = (ab)m 1 EXAMPLE Simplify the expression, x2y 3 xy ÷ 5. z z Solution The expression becomes x2 y 3 z 5 × = xy 2z 4. z xy (b) Negative Integer Indices Law No.