On Logarithmic Functions

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

Properties of logarithmic functions. 3 / 10 log6(6) = 1

Properties of logarithmic functions. 3 / 10 1

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. Let a = log5(x) so that x = 5 5 . Compute

y a y log5(x ) = log5 ((5 ) ) The exponential of an exponential is? =

Properties of logarithmic functions. 6 / 10 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 ) ) The exponential of an exponential is? a·y = log5 (5 )

Properties of logarithmic functions. 6 / 10 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 ) ) The exponential of an exponential is? a·y = log5 (5 ) Exponential and logarithm cancel =

Properties of logarithmic functions. 6 / 10 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 ) ) The exponential of an exponential is? a·y = log5 (5 ) Exponential and logarithm cancel = a · y =

Properties of logarithmic functions. 6 / 10 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 ) ) The exponential of an exponential is? a·y = log5 (5 ) Exponential and logarithm cancel = a · y = log5(x) · y

Properties of logarithmic functions. 6 / 10 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 ) ) 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)

Properties of logarithmic functions. 6 / 10 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 ) ) 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 )

Properties of logarithmic functions. 6 / 10 The diﬀerence of logs is the log of the quotient x2+1 log3 x = 2. exponentiate to get rid of the log x2+1 2 x = 3 = 9. Multiply both sides by x to clear the denominator x2 + 1 = 9x. regroup. x2 − 9x + 1 = 0 By the quadratic√ formula 9 ± 92 − 4 x = . 2

Using these rules. 2 Solve log3(x + 1) − log3(x) = 2.

Properties of logarithmic functions. 7 / 10 x2+1 log3 x = 2. exponentiate to get rid of the log x2+1 2 x = 3 = 9. Multiply both sides by x to clear the denominator x2 + 1 = 9x. regroup. x2 − 9x + 1 = 0 By the quadratic√ formula 9 ± 92 − 4 x = . 2

Using these rules. 2 Solve log3(x + 1) − log3(x) = 2. The diﬀerence of logs is the log of the quotient

Properties of logarithmic functions. 7 / 10 exponentiate to get rid of the log x2+1 2 x = 3 = 9. Multiply both sides by x to clear the denominator x2 + 1 = 9x. regroup. x2 − 9x + 1 = 0 By the quadratic√ formula 9 ± 92 − 4 x = . 2

Using these rules. 2 Solve log3(x + 1) − log3(x) = 2. The diﬀerence of logs is the log of the quotient x2+1 log3 x = 2.

Properties of logarithmic functions. 7 / 10 x2+1 2 x = 3 = 9. Multiply both sides by x to clear the denominator x2 + 1 = 9x. regroup. x2 − 9x + 1 = 0 By the quadratic√ formula 9 ± 92 − 4 x = . 2

Using these rules. 2 Solve log3(x + 1) − log3(x) = 2. The diﬀerence of logs is the log of the quotient x2+1 log3 x = 2. exponentiate to get rid of the log

Properties of logarithmic functions. 7 / 10 9. Multiply both sides by x to clear the denominator x2 + 1 = 9x. regroup. x2 − 9x + 1 = 0 By the quadratic√ formula 9 ± 92 − 4 x = . 2

Using these rules. 2 Solve log3(x + 1) − log3(x) = 2. The diﬀerence of logs is the log of the quotient x2+1 log3 x = 2. exponentiate to get rid of the log x2+1 2 x = 3 =

Properties of logarithmic functions. 7 / 10 Multiply both sides by x to clear the denominator x2 + 1 = 9x. regroup. x2 − 9x + 1 = 0 By the quadratic√ formula 9 ± 92 − 4 x = . 2

Using these rules. 2 Solve log3(x + 1) − log3(x) = 2. The diﬀerence of logs is the log of the quotient x2+1 log3 x = 2. exponentiate to get rid of the log x2+1 2 x = 3 = 9.

Properties of logarithmic functions. 7 / 10 x2 + 1 = 9x. regroup. x2 − 9x + 1 = 0 By the quadratic√ formula 9 ± 92 − 4 x = . 2

Using these rules. 2 Solve log3(x + 1) − log3(x) = 2. The diﬀerence of logs is the log of the quotient x2+1 log3 x = 2. exponentiate to get rid of the log x2+1 2 x = 3 = 9. Multiply both sides by x to clear the denominator

Properties of logarithmic functions. 7 / 10 regroup. x2 − 9x + 1 = 0 By the quadratic√ formula 9 ± 92 − 4 x = . 2

Properties of logarithmic functions. 7 / 10 x2 − 9x + 1 = 0 By the quadratic√ formula 9 ± 92 − 4 x = . 2

Properties of logarithmic functions. 7 / 10 By the quadratic√ formula 9 ± 92 − 4 x = . 2

Properties of logarithmic functions. 7 / 10 Using these rules. 2 Solve log3(x + 1) − log3(x) = 2. The diﬀerence of logs is the log of the quotient x2+1 log3 x = 2. exponentiate to get rid of the log x2+1 2 x = 3 = 9. Multiply both sides by x to clear the denominator x2 + 1 = 9x. regroup. x2 − 9x + 1 = 0 By the quadratic√ formula 9 ± 92 − 4 x = . 2

Properties of logarithmic functions. 7 / 10 For you: 2 Solve log2(x ) − log2(x + 1) = 1. The diﬀerence of logs is the log of the quotient

Properties of logarithmic functions. 8 / 10 z In exponential form z = log2(6) means that 6 = 2 .

Take log3 of each side to see

z log3(6) = log3(2 ) = z · log3(2)

Solving for z, z = log3(6) log3(2) Thus log (6) = log3(6) 2 log3(2) In general this is true. Theorem

logb(x) For any bases a, b > 0 a, b 6= 0 and any x > 0, loga(x) = logb(a) If you understand one Logarithmic function, you understand them all!

Translating between bases

Express z = log2(6) in terms of log3’s.

Properties of logarithmic functions. 9 / 10 Take log3 of each side to see

z log3(6) = log3(2 ) = z · log3(2)

Solving for z, z = log3(6) log3(2) Thus log (6) = log3(6) 2 log3(2) In general this is true. Theorem

logb(x) For any bases a, b > 0 a, b 6= 0 and any x > 0, loga(x) = logb(a) If you understand one Logarithmic function, you understand them all!

Translating between bases

Express z = log2(6) in terms of log3’s. z In exponential form z = log2(6) means that 6 = 2 .

Properties of logarithmic functions. 9 / 10 z · log3(2)

Solving for z, z = log3(6) log3(2) Thus log (6) = log3(6) 2 log3(2) In general this is true. Theorem

logb(x) For any bases a, b > 0 a, b 6= 0 and any x > 0, loga(x) = logb(a) If you understand one Logarithmic function, you understand them all!

Translating between bases

Express z = log2(6) in terms of log3’s. z In exponential form z = log2(6) means that 6 = 2 .

Take log3 of each side to see

z log3(6) = log3(2 ) =

Properties of logarithmic functions. 9 / 10 Solving for z, z = log3(6) log3(2) Thus log (6) = log3(6) 2 log3(2) In general this is true. Theorem

logb(x) For any bases a, b > 0 a, b 6= 0 and any x > 0, loga(x) = logb(a) If you understand one Logarithmic function, you understand them all!

Translating between bases

Express z = log2(6) in terms of log3’s. z In exponential form z = log2(6) means that 6 = 2 .

Take log3 of each side to see

z log3(6) = log3(2 ) = z · log3(2)

Properties of logarithmic functions. 9 / 10 z = log3(6) log3(2) Thus log (6) = log3(6) 2 log3(2) In general this is true. Theorem

logb(x) For any bases a, b > 0 a, b 6= 0 and any x > 0, loga(x) = logb(a) If you understand one Logarithmic function, you understand them all!

Translating between bases

Express z = log2(6) in terms of log3’s. z In exponential form z = log2(6) means that 6 = 2 .

Take log3 of each side to see

z log3(6) = log3(2 ) = z · log3(2)

Solving for z,

Properties of logarithmic functions. 9 / 10 Thus log (6) = log3(6) 2 log3(2) In general this is true. Theorem

logb(x) For any bases a, b > 0 a, b 6= 0 and any x > 0, loga(x) = logb(a) If you understand one Logarithmic function, you understand them all!

Translating between bases

Express z = log2(6) in terms of log3’s. z In exponential form z = log2(6) means that 6 = 2 .

Take log3 of each side to see

z log3(6) = log3(2 ) = z · log3(2)

Solving for z, z = log3(6) log3(2)

Properties of logarithmic functions. 9 / 10 In general this is true. Theorem

logb(x) For any bases a, b > 0 a, b 6= 0 and any x > 0, loga(x) = logb(a) If you understand one Logarithmic function, you understand them all!

Translating between bases

Express z = log2(6) in terms of log3’s. z In exponential form z = log2(6) means that 6 = 2 .

Take log3 of each side to see

z log3(6) = log3(2 ) = z · log3(2)

Solving for z, z = log3(6) log3(2) Thus log (6) = log3(6) 2 log3(2)

Properties of logarithmic functions. 9 / 10 If you understand one Logarithmic function, you understand them all!

Translating between bases

Express z = log2(6) in terms of log3’s. z In exponential form z = log2(6) means that 6 = 2 .

Take log3 of each side to see

z log3(6) = log3(2 ) = z · log3(2)

Solving for z, z = log3(6) log3(2) Thus log (6) = log3(6) 2 log3(2) In general this is true. Theorem

logb(x) For any bases a, b > 0 a, b 6= 0 and any x > 0, loga(x) = logb(a)

Properties of logarithmic functions. 9 / 10 Translating between bases

Express z = log2(6) in terms of log3’s. z In exponential form z = log2(6) means that 6 = 2 .

Take log3 of each side to see

z log3(6) = log3(2 ) = z · log3(2)

Solving for z, z = log3(6) log3(2) Thus log (6) = log3(6) 2 log3(2) In general this is true. Theorem

logb(x) For any bases a, b > 0 a, b 6= 0 and any x > 0, loga(x) = logb(a) If you understand one Logarithmic function, you understand them all!

Properties of logarithmic functions. 9 / 10 Closing example

Express log4(3) in terms of log3’s to simplify

log3(4) · log4(3)

Properties of logarithmic functions. 10 / 10