Further Logarithms - Mathematics Form 3 Notes

Introduction

If y = athen we introduce the inverse function logarithm and define logy=x (Read as log base a of  y equals x).

In general

y = ax ⇔ logy=x

Where ⇔means “implies and is implied by” i.e. it works both ways!

Note this means that, going from exponent form to logarithmic form:

 102 = 100 ⇒ log10 (100) = 2 10-1=0.01 ⇒ log10 (0.01) = -2 100 = 1 ⇒ log10 (1) = 0 25 =32 ⇒ log2 (32) = 5 9½ = 3 ⇒ log9 (3) = ½ 8⅔ = 4 ⇒ log8 (4) = ⅔

And in going from logarithmic form to exponent form:

 log10 (10) = 1 ⇒ 101 = 10 log10 (0.001) = -3 ⇒ 10-3=0.001 log2 (1) = 0 ⇒ 20 = 1 log3 (81) = 4  ⇒ 34 = 81 log100 (10) = ½ ⇒ 100½ =10 log5 (5√5) = 3/2 ⇒ 53/2 = 5√5

Laws of Logarithms

Product and Quotient Laws of Logarithms

The Product Law

loga (M×N) = loga M + loga N

The Quotient Law

log a (M/N) = loga M − loga N

Example.

log6 9 + log8 − log6 2

= log6 72 − log6 2
= log6 (72/2) = log36
= 2

The Power Law of Logarithms

Loga Mn = nloga M

Example.

2log 5 + 2log 2

= log 52 + log 22

= log 25 + log 4

= log 100 = log10100

= 2

Logarithm of a Root

logb x1/n = 1/n logb x  or  logb n√x = logx
n

Example.

log3 5√27 ⇒ log3 271/5 ⇒ 1/5log3 27 = 1/5(3) = 3/5

Proof of Properties

 Property Proof Reason for Step 1. logb b = 1 and logb 1 = 0 b1 = b and b0 = 1 Definition of logarithms 2. (product rule)logb xy = logb x + logb y a. Let logb x = m and logb y = nb. x = bm and y = bnc. xy = bm × bnd. xy = bm + ne. logb xy = m + nf. logb xy = logb x + logb y a. Setupb. Rewrite in exponent formc. Multiply togetherd. Product rule for exponentse. Rewrite in log formf. Substitution 3. (quotient rule)logb x/y = logb x − logb y a. Let logb x = m and logb y = nb. x = bm and y = bnc. x/y = bm            bnd. x = bm−n    ye. logb x/y = m − nf. logb x/y = logb x − logb y a. Given: compact formb. Rewrite in exponent formc. Divided. Quotient rule for exponentse. Rewrite in log formf. Substitution 4. (power rule)logb xn = nlogb x a. Let m = logb x so x = bmb. xn = bmnc. logb xn = mnd. logb xn = nlogb x a. Setupb. Raise both sides to the nth powerc. Rewrite as logd. Substitute 5. Properties used to solve log equations:a. if bx = by, then x = yb. if logb x = logb y, then x= y a. This follows directly from the properties for exponents.b. i. logb x - logb y = 0ii. logb x/y = 0iii. x/y=b0iv. x/y =1 so x = y b. i. Subtract from both sidesii. Quotient ruleiii. Rewrite in exponent formiv. b0 = 1

Solving Exponential and Logarithmic Equations

By taking logarithms, and exponential equation can be converted to a linear equation and solved.

We will use the process of taking logarithms of both sides.

Example.

4x = 12

log 4x = log 12

xlog 4 = log 12

x = log 12
log 4

x= 1.792

Note;

• A logarithmic expression is defined only for positive values of the argument.
• When we solve a logarithmic equation it is essential to verify that the solution(s) does not result in the logarithm of a negative number.
• Solutions that would result in the logarithm of a negative number are called extraneous, and are not valid solutions.

Example.

Solve for x:

log(x+1) + log5 (x − 3) = 1 → (the 1 is the same as log5)

log(x+1)(x − 3) = log5

(x+1)(x − 3) = 5

x2 − 2x − 3 − 5 = 0

x2 − 2x − 8 = 0

(x−4)(x+2) = 0

x = 4, x= −2(extraneous)

Verify:

log5 (4+1) + log5 (4 − 3) = 1
log5 + log1 = 1
1+0=1

log5 (-2+1) + log5 (-2 − 3) = 1
log-1 + log-5 not possible

Solving Equations Using Logs

Examples

1. Solve the equation 10x =3.79

The definition of logs says if If y = ax then logy=x or y = a⇔ logy=x
Hence 10x = 3.79 ⇒ xc= log10 3.79 = 0.57864 (to 5 decimal places)
Check 100.57864 = 3.79000 (to 5 decimal places)

In practice from we take logs to base 10 giving
log10 10x = log10 3.79
xlog10 10 = log10 3.79
x = 0.57864
2. Solve the equation 32x = 56
log10 32x = log10 56
2xlog10 3 = log10 56
2x = log10 56 = 3.66403
log10 3
x= 1.83201
Check 33 = 27, 34 = 81, we want 32x so the value of 2x lies between 3 and 4 or 3<2x<4 which means x lies between 1.5 and 2. This tells us that x= 1.83201 is roughly correct.
3. Solve the equation 4x = 3x+1
x log10 4 = (x+1)log10 3
x log10 4 = xlog10 3 +  log10 3
x log10 4 − xlog10 3 = log10 3
x(log10 4 − log10 3) = log10 3
x =         log10 3          = 3.8188
log10 4 − log10 3

Check 4x = 43.8188 ≈ 44 = 256 and 3x+1 = 34.8188 ≈ 35 = 243 are very close!

Note you could combine terms, giving,
x =         log10 3          =  log10 3  = 3.8188
log10 4 − log10 3      log10 4/3
4. Solve the equation
4x+6 = 35 − 2x

Take logs of both sides            log 4x+6 = log 35 − 2x
(x+6)log 4 = (5 − 2x)log 3
Expand brackets                     xlog 4 + 6log 4 = 5log 3 − 2xlog 3
Collect terms                          xlog 4 + 2xlog 3 = 5log 3 − 6log 4
Factorise the left hand side and divide    x=  5log 3 − 6log 4 =   −0.78825
log 4 + 2log 3
(Note you get the same answer by using the ln button on your calculator.)

Check 4x+6 = 4−0.78825+6 = 45.21175 = 1373.368 and 35 −2(−0.78825) = 36.576498 = 1373.368

Notice that you could combine the log-terms in 5log 3 − 6log 4
to give log (35 ÷ 46)
log 4 + 2log 3             log (4 × 32)

It does not really simplify things here but, in some cases, it can.
5. Solve the equation
7(3x−1) = 2(52x+1)

Take logs of both sides                      log 7 + (x − 1)log 3 = log 2+ (2x+1)log 5
Expand brackets                               log 7 + xlog 3 − log 3 = log 2+ 2xlog 5 + log 5
Collect terms                                    xlog 3 − 2xlog 5 = log 2 + log 5 + log 3 − log 7
Factorize left hand side                      x(log 3 − 2 log 5) = log 2 + log 5 + log 3 − log 7
simplify                                            xlog (3/52) = log ((2 × 5 × 3)/7)
xlog (3/25) = log (30/7)
divide                                               x = log (30/7) = −0.686371
log (3/25)
Check

LHS = 7(3x−1) ≈ 7(3−0.686371−1) ≈ 7×3−1.7 ≈ 7/32 ≈ 7/(taking 31.7 ≈ 32 = 9)
RHS = 2(52x+1) = 2(52×−0.686371+1) ≈ 2 × 5−0.42/50.4 = 2/√5 ≈ 1 (taking  50.4 ≈ 50.5 = √5 = 2.2...)

The values of LHS and RHS are roughly the same. A more exact check could be made using a calculator.

Logarithmic Equations and Expressions

Consider the following equations

log381 = x and logx 8 = 3

The value of x in each case is established as follows

log381 = x
Therefore 3x = 81
3x = 34
x =4

log8 = 3
x
3 = 8
x3 =
23
x = 2

Example

Solve log2

Solution

Let log62 = t. then 6t= 2

Introducing logarithm to base 1 0 on both sides

log 6t = log 2
t log 6 = log 2
t = log 2
log 6
t = 0.3010
0.7782
t = 0.3868
Therefore log62 = 0.3868

Example

22x + 3(2x) − 4 = 0

Taking logs on both sides cannot help in getting the value of x, since 22x + 3(2x) cannot be combined into a single expression.

However if we let 2x = y then the equation becomes quadratic in y.

Solution

Thus, let 2x = y…………….. (1 )
Therefore
y2 + 3y − 4 = 0…………………(2)
(y + 4)(y - 1) = 0

y = − 4 or y = 1
Substituting for y in equation (1 );
Let
2x = -4 or let 2x = 1

There is no real value of x for which 2x = -4 hence 2x = 1= 2thus x = 0

Example

Solve for x in (log10x)2 = 3 − log10x2

Solution

Let log10x = t……………………….(1)
Therefore t2 = 3 − 2t
t
2 + 2 t − 3 = 0 solve the quadratic equation using any method
t2 + 3t − t - 3 = 0
t(t + 3) - 1(t − 3) = 0
(t − 1)(t + 3) = 0
t = 1 or t = − 3
Substituting for t in the equation (1 ).
log10x = 1 or log10x = −3
10
1 = x or 10-3= x
x = 10 or 1/1000

Note;

log    1
log
a b

Past KCSE Questions on the Topic

1. Solve for (log3x)2 – ½ log3x = 3/2
2. Find the values of x which satisfy the equation 52x – 6 (5x) + 5 =0
3. Solve the equation
Log (x + 24) – 2 log 3 = log (9 – 2x)
4. Find the value of x in the following equation 49(x+1) + 7(2x) = 350
5. Find x if 3 log 5 + log x2 = log 1/125
6. Without using logarithm tables, find the value of x in the equation
Log x3 + log 5x = 5 log2 – log 2/5
7.
1. Given that P = 3y express the questions 3(2y -1) + 2 x 3(y-1) = 1 in terms of P
2. Hence or otherwise find the value of y in the equation: 3(2y -1) + 2 x 3(y-1) = 1

• ✔ To read offline at any time.
• ✔ To Print at your convenience
• ✔ Share Easily with Friends / Students

Related items

.
Subscribe now

access all the content at an affordable rate
or
Buy any individual paper or notes as a pdf via MPESA
and get it sent to you via WhatsApp