1a 2009 Logarithm

Logarithms In this chapter you must be able to (i) understand the relationship between logarithms and indices, and use the laws of logarithms (excluding change of base) (ii) understand the definition and properties of ex and ln x, including the relationship as inverse functions and their graphs Relationship between log and indices Let x = a p ⇔ loga x = p ; y = a q ⇔ loga y = q xy = a p . a q = a p + q ; log a xy = log a x + log a y x ap = = a p−q y aq

;

log a

x = log a x − log a y y

x n = (ap )n = ap n = log a x log a a = 1 ; log a 1 = 0 i) Express log

x3

n

= n log a x

in terms of log x and log y.

y ii) Express 3 log 8 − log 2 5 as a single logarithm. iii) Solve the equation log 6 x + log 6 ( x + 5) = 2 . 1

(3logx − 12 logy)

(log5) (x = 4; x = −9(not acceptable))

Graphs of ex and ln x

1

y y

y = ln x

y=ex

x

Note these two: (1) e ln x = x 1 Solve for x i) ex - 2 - 3e-x = 0 ii) e 3t + 2et = 3e2t

(2) ln e x = x ( x = ln 3) (t = 0, ln 2)

2

−b

iii) - 2 ln x = b

(x = e 2 ) (x = e2 )

iv) 3 ln x – ½ ln x = 5 v) ln x = 2 ( ln 3 - ln 5 ) vi) ak – ekx = 0 vii)x ln x = e

( x = 9 25)

( x = ln a ) ( x = e, e −1 )

viii) 2 + 5 e − 4x = 3 ix) - ln x = x) ln x =

1 3

t 50

+c ( x = 4)

(ln 16 + 2 ln 2)

xi) ax = e 2x +1 xii) 2 (ln 1 ( x = ± 2 ae ) xiii)32x = 42 – x xiv)ex - 2e— x = 1

2x)

=

1

+

ln

a

2 Solve the following equations giving x correct to 3 significant figures. ( x = 0.642) a) 7x . 8 2x - 1 = 6.3 b) c) d) e)

32x – 3x +2 + 8 = 0 2x = 3 3x+1 = 18 2x . 32x = 5

( x = 0 ; 1.893)

( 0.5568)

3 Solve for x a) ex ( 1 - 2x ) = 0 ( x = 12) b) e-x ( 1 + x ) = 0 {e

c) d)

−x

= 0 (not possible); x = −1} ( dont cancel;

xex = ex ex - e-x = 6

x = 1)

(1.818)

4 By using a suitable substitution, solve the following equations: a) 2 2x - 5 . 2x + 6 = 0 b) 32x + 1 + 32 = 3x + 3 + 3x 5

(x =

ln 3 ln 2

= 1.5849; x = 1)

( x = −1; x = 2)

If y = ln ( 1 + 6x + 8x2 ) - ln ( 1 + 5x + 4x2 ) + ln ( 1+ x )- ln (1 + 2x ), prove that y is independent of x.

6 Solve for x and y : a) x - y = 1 ; 2x . 3y = 432

3

b) log x + log y = log 10 ; log x - log y = log 2.5 c) 2x . 4- y = 2 ; 3- x . 92y = 3 d) log x + log y = 1 ; log x - log y = log 2.5 e) x1/2 + y 2/3 = 6 ; x

- 1/2

+y

- 2/3

=¾

f) log 2 (xy 2) = 4 ; log 2 x . log 2 y = 2

( x = 3, y = 1) ( x = 5, y = 2) 3

( x = 16, y = 2 2 ; x = 4, y = 8) ( x = 4, y = 2)

7) log 4 2 . log 4 ( 1 2 x ) = log 4 x 8) Express ln (2√ e) - 1/3 ln (8/e) - ln (e/3) in the form c + ln d. (− 1 6 + ln 3)

4