Properties of logarithmic functions. Properties of logarithmic functions. 1 / 10 What today's all about Today we develop four important formula involving logarithms. I'll give myself board space for all four right now. The Logarithm of a product logb(x · y) = ??? The Logarithm of a quotient logb(x=y) = ??? y The Logarithm of an exponent logb(x ) = ??? The Logarithm with respect to a different base logc (x) = ??? Properties of logarithmic functions. 2 / 10 = log(35+2) But logarithm undoes exponentiation. 5 2 log3(3 · 3 ) = 5 + 2. 5 2 On the other hand log3(3 ) + log3(3 ) = 5 + 2 There is a general principle at work here: Theorem For any base b > 0, b 6= 1, and any x > 0, y > 0 logb(x · y) = logb(x) + logb(y) Compute: log6(2) + log6(3) = log6(2 · 3) = log6(6) = 1 The logarithm of a product Compute 5 2 log3(3 · 3 ) Properties of logarithmic functions. 3 / 10 But logarithm undoes exponentiation. 5 2 log3(3 · 3 ) = 5 + 2. 5 2 On the other hand log3(3 ) + log3(3 ) = 5 + 2 There is a general principle at work here: Theorem For any base b > 0, b 6= 1, and any x > 0, y > 0 logb(x · y) = logb(x) + logb(y) Compute: log6(2) + log6(3) = log6(2 · 3) = log6(6) = 1 The logarithm of a product Compute 5 2 5+2 log3(3 · 3 ) = log(3 ) Properties of logarithmic functions. 3 / 10 5 + 2. 5 2 On the other hand log3(3 ) + log3(3 ) = 5 + 2 There is a general principle at work here: Theorem For any base b > 0, b 6= 1, and any x > 0, y > 0 logb(x · y) = logb(x) + logb(y) Compute: log6(2) + log6(3) = log6(2 · 3) = log6(6) = 1 The logarithm of a product Compute 5 2 5+2 log3(3 · 3 ) = log(3 ) But logarithm undoes exponentiation. 5 2 log3(3 · 3 ) = Properties of logarithmic functions. 3 / 10 5 2 On the other hand log3(3 ) + log3(3 ) = 5 + 2 There is a general principle at work here: Theorem For any base b > 0, b 6= 1, and any x > 0, y > 0 logb(x · y) = logb(x) + logb(y) Compute: log6(2) + log6(3) = log6(2 · 3) = log6(6) = 1 The logarithm of a product Compute 5 2 5+2 log3(3 · 3 ) = log(3 ) But logarithm undoes exponentiation. 5 2 log3(3 · 3 ) = 5 + 2. Properties of logarithmic functions. 3 / 10 5 + 2 There is a general principle at work here: Theorem For any base b > 0, b 6= 1, and any x > 0, y > 0 logb(x · y) = logb(x) + logb(y) Compute: log6(2) + log6(3) = log6(2 · 3) = log6(6) = 1 The logarithm of a product Compute 5 2 5+2 log3(3 · 3 ) = log(3 ) But logarithm undoes exponentiation. 5 2 log3(3 · 3 ) = 5 + 2. 5 2 On the other hand log3(3 ) + log3(3 ) = Properties of logarithmic functions. 3 / 10 Compute: log6(2) + log6(3) = log6(2 · 3) = log6(6) = 1 The logarithm of a product Compute 5 2 5+2 log3(3 · 3 ) = log(3 ) But logarithm undoes exponentiation. 5 2 log3(3 · 3 ) = 5 + 2. 5 2 On the other hand log3(3 ) + log3(3 ) = 5 + 2 There is a general principle at work here: Theorem For any base b > 0, b 6= 1, and any x > 0, y > 0 logb(x · y) = logb(x) + logb(y) Properties of logarithmic functions. 3 / 10 log6(2 · 3) = log6(6) = 1 The logarithm of a product Compute 5 2 5+2 log3(3 · 3 ) = log(3 ) But logarithm undoes exponentiation. 5 2 log3(3 · 3 ) = 5 + 2. 5 2 On the other hand log3(3 ) + log3(3 ) = 5 + 2 There is a general principle at work here: Theorem For any base b > 0, b 6= 1, and any x > 0, y > 0 logb(x · y) = logb(x) + logb(y) Compute: log6(2) + log6(3) = Properties of logarithmic functions. 3 / 10 log6(6) = 1 The logarithm of a product Compute 5 2 5+2 log3(3 · 3 ) = log(3 ) But logarithm undoes exponentiation. 5 2 log3(3 · 3 ) = 5 + 2. 5 2 On the other hand log3(3 ) + log3(3 ) = 5 + 2 There is a general principle at work here: Theorem For any base b > 0, b 6= 1, and any x > 0, y > 0 logb(x · y) = logb(x) + logb(y) Compute: log6(2) + log6(3) = log6(2 · 3) = Properties of logarithmic functions. 3 / 10 1 The logarithm of a product Compute 5 2 5+2 log3(3 · 3 ) = log(3 ) But logarithm undoes exponentiation. 5 2 log3(3 · 3 ) = 5 + 2. 5 2 On the other hand log3(3 ) + log3(3 ) = 5 + 2 There is a general principle at work here: Theorem For any base b > 0, b 6= 1, and any x > 0, y > 0 logb(x · y) = logb(x) + logb(y) Compute: log6(2) + log6(3) = log6(2 · 3) = log6(6) = Properties of logarithmic functions. 3 / 10 The logarithm of a product Compute 5 2 5+2 log3(3 · 3 ) = log(3 ) But logarithm undoes exponentiation. 5 2 log3(3 · 3 ) = 5 + 2. 5 2 On the other hand log3(3 ) + log3(3 ) = 5 + 2 There is a general principle at work here: Theorem For any base b > 0, b 6= 1, and any x > 0, y > 0 logb(x · y) = logb(x) + logb(y) Compute: log6(2) + log6(3) = log6(2 · 3) = log6(6) = 1 Properties of logarithmic functions. 3 / 10 3a log 3 3b Quotients of exponentials are well behaved 3a log = log 3a−b 3 3b 3 Exponentiation undoes logarithm x log = log 3a−b = a − b = log (x) − log (y) 3 y 3 3 3 The Logarithm of a quotient a Consider any x and y. Let a = log3(x) and b = log3(y) so that x = 3 and y = 3b Make this substitution to compute x log = 3 y Properties of logarithmic functions. 4 / 10 Quotients of exponentials are well behaved 3a log = log 3a−b 3 3b 3 Exponentiation undoes logarithm x log = log 3a−b = a − b = log (x) − log (y) 3 y 3 3 3 The Logarithm of a quotient a Consider any x and y. Let a = log3(x) and b = log3(y) so that x = 3 and y = 3b Make this substitution to compute x 3a log = log 3 y 3 3b Properties of logarithmic functions. 4 / 10 a−b log3 3 Exponentiation undoes logarithm x log = log 3a−b = a − b = log (x) − log (y) 3 y 3 3 3 The Logarithm of a quotient a Consider any x and y. Let a = log3(x) and b = log3(y) so that x = 3 and y = 3b Make this substitution to compute x 3a log = log 3 y 3 3b Quotients of exponentials are well behaved 3a log = 3 3b Properties of logarithmic functions. 4 / 10 Exponentiation undoes logarithm x log = log 3a−b = a − b = log (x) − log (y) 3 y 3 3 3 The Logarithm of a quotient a Consider any x and y. Let a = log3(x) and b = log3(y) so that x = 3 and y = 3b Make this substitution to compute x 3a log = log 3 y 3 3b Quotients of exponentials are well behaved 3a log = log 3a−b 3 3b 3 Properties of logarithmic functions. 4 / 10 log3(x) − log3(y) The Logarithm of a quotient a Consider any x and y. Let a = log3(x) and b = log3(y) so that x = 3 and y = 3b Make this substitution to compute x 3a log = log 3 y 3 3b Quotients of exponentials are well behaved 3a log = log 3a−b 3 3b 3 Exponentiation undoes logarithm x log = log 3a−b = a − b = 3 y 3 Properties of logarithmic functions. 4 / 10 The Logarithm of a quotient a Consider any x and y. Let a = log3(x) and b = log3(y) so that x = 3 and y = 3b Make this substitution to compute x 3a log = log 3 y 3 3b Quotients of exponentials are well behaved 3a log = log 3a−b 3 3b 3 Exponentiation undoes logarithm x log = log 3a−b = a − b = log (x) − log (y) 3 y 3 3 3 Properties of logarithmic functions. 4 / 10 Compute log10(50) − log10(5) The logarithm of a quotient is the difference of the logarithms We've proven a theorem: Theorem For any base b > 0, b 6= 1, and any x > 0, y > 0 logb(x=y) = logb(x) − logb(y) Properties of logarithmic functions. 5 / 10 The logarithm of a quotient is the difference of the logarithms We've proven a theorem: Theorem For any base b > 0, b 6= 1, and any x > 0, y > 0 logb(x=y) = logb(x) − logb(y) Compute log10(50) − log10(5) Properties of logarithmic functions. 5 / 10 The exponential of an exponential is? a·y = log5 (5 ) Exponential and logarithm cancel = a · y = log5(x) · y Theorem For any base b > 0 b 6= 1 and any x > 0, and y y logb(x ) = y · logb(x) 1 − log (3) Compute log2(3 2 ) Discuss in your groups: Logarithms of powers log (a) Consider any x > 0 and y. Let a = log5(x) so that x = 5 5 . Compute y a y log5(x ) = log5 ((5 ) ) Properties of logarithmic functions. 6 / 10 a·y log5 (5 ) Exponential and logarithm cancel = a · y = log5(x) · y Theorem For any base b > 0 b 6= 1 and any x > 0, and y y logb(x ) = y · logb(x) 1 − log (3) Compute log2(3 2 ) Discuss in your groups: Logarithms of powers log (a) Consider any x > 0 and y.