LOGARITHMS

Definition:

The “Log” of a number, to a given base, is the power to whichthe base must be raised in order to equal the number.

e.g.your calculator tells you that log 100= 2

This is because the log button on the calculator uses base 10.

…and 10 must be raised to the power 2, to equal 100.

i.e 102 = 100

We write: log10100 = 2 ( Since 102 = 100 )

n.b. log to base 10 is sometimes written as lg i.e. lg100 = 2

Note that any positive value can be chosen as the base number:

e.g. In base 2; log28 = 3 ( Since 23 = 8 )

In general:

If loga x = p , then ap = x

e.g. In base 7; log749 = 2 ( Since 72 = 49 )

Note also: log33 = 1

log55 = 1i.e. For any base: logaa = 1

Also: log31 = 0

log81 = 0i.e. For any base: loga1 = 0

Example 1: Find the value of: a) log232 b) log164 c) log2 81

a) log232 = 5 ( Since 25 = 32 )

b) log164 =

Example 2: Solve the following equations: a) log4x = 2 b) log3(2x + 1) = 2

a) log4x = 2 Hence, x = 16

b) log3(2x + 1) = 2

9 = 2x + 1 Hence, x = 4

( Since 16 ½ = 16 = 4 )

c) log218 = –3 ( Since 2

–3 = 123

18= )

42 = x

32 = 2x + 1

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Laws of Logarithms

As logarithms are themselves indices the laws of logs are closelyrelated to the laws of indices.

If logaX = p, then ap = X. If logaY = q, then aq = Y

i) XY = ap aq i.e XY = ap + q So, logaXY = p + q

logaXY = logaX + logaY

Ylog Xlog YXlog aaa

iii) Xn = (ap)n i.e. Xn = apn So, logaXn = pn

logaXn = nlogaX

So, log aXY = p – q ii) X

Yap

aq= a p – q X

Y =i.e.

Example 1: Express the following as a single logarithm: a) log4 + 2log3 b) logP – 2logQ + 3logR

a) log4 + 2log3 = log4 + log32 = log 36

b) logP – 2logQ + 3logR = logP – logQ2 + logR3

Example 2: Express log in terms of log x and log y. xy2

= log PR3

Q2

log xy2 = log x – log y2 1

2= log x – 2 log y

Example 3: Given log23 = p, and log25 = q, find the following in terms of p and q: a) log215 b) log260 c) log22.7.

a) log215 = log2(3)(5) = log23 + log25 = p + q

b) log260 = log2(3)(5)(22) = log23 + log25 + log222

= log23 + log25 + 2log22

= p + q + 2

c) log22.7 = = log227 – log210 = log233 – log2(2)(5)

= 3log23 – ( log22 + log25 )

= 3p –1 – q

or justlog24

log22710

Example 4: Solve log4x + log4(3x + 2) = 2.

(3x + 8)(x – 2) = 0

Since x must be positive,

log4 x (3x + 2) = 2

3x2 + 2x – 16 = 0

Using: logX + logY = logXY

x = 2

42 = x (3x + 2 )

Equations of the form ax = b.

These can be solved using logarithms:

Example 5: Solve the equation 2 = 7.x

Taking logs of both sides:

log 2 = log 7x

logXn = nlogXUsing:

x log 2 = log 7

x = log 7log 2

= 2.81 ( 3 sig.figs.)

Summary of key points:

This PowerPoint produced by R.Collins ; Updated Feb.2012

The “Log” of a number, to a given base, is the power to whichthe base must be raised in order to equal the number.

If loga x = p, then ap = x

logaa = 1 loga1 = 0

Laws of Logs: logX + logY = logXY

YXlog logY logX

logXn = n logX