Cambridge Additional Mathematics

GRAPHING

PACKAGE

Logarithms (Chapter 5) 147

Example 23 Self Tutor

Find algebraically the exact points of intersection of y=ex¡ 3 and y=1¡ 3 e¡x.

The functions meet where

ex¡3=1¡ 3 e¡x

) ex¡4+3e¡x=0

) e^2 x¡ 4 ex+3=0 fmultiplying each term byexg

) (ex¡1)(ex¡3) = 0

) ex=1or 3

) x=ln1or ln 3

) x=0or ln 3

When x=0, y=e^0 ¡3=¡ 2

When x=ln3, y=eln 3¡3=0

) the functions meet at (0,¡2) and at (ln 3,0).

9 Solve forx:

a e^2 x=2ex b ex=e¡x c e^2 x¡ 5 ex+6=0

d ex+2=3e¡x e 1+12e¡x=ex f ex+e¡x=3

10 Find algebraically the point(s) of intersection of:

a y=ex and y=e^2 x¡ 6 b y=2ex+1 and y=7¡ex

c y=3¡ex and y=5e¡x¡ 3

A logarithm in basebcan be written with a different basecusing thechange of base rule:

logba=

logca

logcb

for a,b,c> 0 and b,c 6 =1.

Proof: If logba=x, then bx=a

) logcbx= logca ftaking logarithms in basecg

) xlogcb= logca fpower law of logarithmsg

) x=

logca

logcb

) logba=

logca

logcb

We can use this rule to write logarithms in base 10 or basee. This is useful in helping us evaluate them on

our calculator.

G The change of base rule

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Y:\HAESE\CAM4037\CamAdd_05\147CamAdd_05.cdr Tuesday, 21 January 2014 2:49:58 PM BRIAN