Indices and laws of logarithms
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46: Indices and Laws of 46: Indices and Laws of Logarithms Logarithms © Christine Crisp “ “ Teach A Level Maths” Teach A Level Maths” Vol. 1: AS Core Vol. 1: AS Core Modules Modules
Transcript of Indices and laws of logarithms
46: Indices and Laws of 46: Indices and Laws of LogarithmsLogarithms
© Christine Crisp
““Teach A Level Maths”Teach A Level Maths”
Vol. 1: AS Core Vol. 1: AS Core ModulesModules
Indices and Laws of
Logarithms
Module C2
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Indices and Laws of
LogarithmsUnknown Indices
Because of important practical applications of growth and decay, we need to be able to solve equations of the type
ba x
where a and b are constants.Equations with unknown indices are solved
using logarithms. We will see what a logarithm is and develop some rules that help us to solve equations.
We have met the graph of and seen that it represents growth or decay.
xay
Indices and Laws of
Logarithms
Ans: If we notice that 3101000
e.g. How would you solve ?100010 x
Indices and Laws of
Logarithms
e.g. How would you solve
Ans: If we notice that 3101000
3 x
31010 x
We can use the same method to solve
813 x or
2552 x
122 xx4 x
433 x 22 55 x
then, (1) becomes
- - - - (1)100010 x
Indices and Laws of
Logarithms
We need to write 75 as a power ( or index ) of 10.
Suppose we want to solve
7510 x
This index is called a logarithm ( or log ) and 10 is the base.
Our calculators give us the value of the logarithm of 75 with a base of 10.
87511010 xThe value is ( 3 d.p. ) so,
8751
8751 x
Tip: It’s useful to notice that, since 75 lies between 10 and 100 ( or ), x lies between 1 and 2.
21 1010 and
The button is markedlog
Indices and Laws of
LogarithmsA logarithm is just an index.To solve an equation where the index is unknown, we can use logarithms.
e.g. Solve the equation giving the answer correct to 3 significant figures.
410 x
x is the logarithm of 4 with a base of 10 4log410 10 xxWe
write
In general if
bx 10 then bx 10log
log
index
( 3 s.f. )6020
Indices and Laws of
Logarithms
( 2 d.p. )
362 x
Solution:
230log 10x
(b) 50log2 10 x
30102 x 150 x ( 2 d.p. )
Exercise
23010 x
Solve the following equations giving the answers correct to 2 d.p.
(a) (b) 50102 x
(a) )32( xthat Notice23010 x
Indices and Laws of
Logarithms
230log23010 10 xx
Generalizing this,
In the exercise, we saw that
This relationship is also true changing from the log form to the index form,
bx 10 bx 10log
Indices and Laws of
Logarithms
bx 10
230log23010 10 xx
Generalizing this,
In the exercise, we saw that
This relationship is also true changing from the log form to the index form,
bx 10log
Indices and Laws of
Logarithms
bx 10
230log23010 10 xx
Generalizing this,
bxbx10log10
This relationship is also true changing from the log form to the index form,
In the exercise we used logs with a base of 10 but the definition holds for any base, so
bxba ax log
so,
bx 10log
Base
Indices and Laws of
Logarithmsba x The
equation
BUT there are no values for logs with base 2 on our calculators so we can’t find this as a
simple number.We need to develop some laws of logs
to enable us to solve a variety of equations with unknown indices or
logs
When the base, a, is 10, we found the equation is easy to solve.e.g. Solve the
equation27510 x
Solution:
27510 x 275log10x) s.f. 3( 442x
52 xe.g. To solve
we could write
5log 2x
Indices and Laws of
Logarithms
2log2 10
2log3 10
30102
30103
4log10
8log10
( from the calculator )30102log10
e.g.
A law of logs for
ka xlog
210 2logAlso,
310 2logAnd,
6020 ( from calculator )
9030 ( from calculator )
Indices and Laws of
Logarithms
2log2 10
2log3 10
310 2log
210 2log
30102
30103
6020
9030
4log10
8log10
( from the calculator )30102log10
e.g.
Also,
And,
( from calculator )
( from calculator )
A law of logs for
ka xlog
Indices and Laws of
Logarithms
2log2 10
2log3 10
310 2log
210 2log
30102
30103
6020
9030
4log10
8log10
( from the calculator )30102log10
e.g.
Also,
And,
( from calculator )
( from calculator )
A law of logs for
ka xlog
We get xkx k 1010 loglog
Indices and Laws of
Logarithms
xkx ak
a loglog
The same reasoning holds for any base, a, so
( the “power to the front ” law of logs )
A law of logs for
ka xlog
Indices and Laws of
Logarithms
Solving ba x
52 xe.g.1 Solve
Solution: 52 x
2log
5log
10
10x
) s.f. ( 3322
5log2log 1010 x
We “take” logs
( Notice that 2 < x < 3 since ) 8242 32 and
5log2log 1010 x
Using the “power to the front” law, we can simplify the l.h.s.
Indices and Laws of
Logarithms
x)3(1001000 e.g.2 Solve the equationSolution: We must change the equation into the form before we take logs.xab
Using the “power to the front” law:
x3log
10log
) s.f. ( 3102 x
x3log10log
x)3(1001000 x310 Divide by 100:
Take logs:
3log10log x
Solving ba x
Indices and Laws of
LogarithmsSUMMARY
bxba ax log
The Definition of a Logarithm
Solving the equation bna x
• “Take” logs
The “Power to the Front” law of logs:
xkx ak
a loglog
• Use the power to the front law• Rearrange to find x.
• Divide by n
Indices and Laws of
LogarithmsExercises
143 x
14log3log 1010 x
( 2 d.p. )
1. Solve the following equations giving the answers correct to 2 d.p.(a) (b) 15122 x
4023log
14log
10
10 x
(a) “Take” logs: 14log3log 1010 x
0898112log
15log2
10
10 x
(b) 15log12log 102
10 x“Take” logs: 15log12log2 1010 x
( 2 d.p. )540 x
Indices and Laws of
Logarithms
2. Solve the equation giving the answer correct to 2 d.p.
x)2(200500
Solution: Divide by 200: xx 252)2(200500
x2log52log Take logs:
Power to the front:
2log52log x
Rearrange: x
2log
52log
( 2 d.p. )321 x
Exercises
Indices and Laws of
Logarithms
