- Introduction
- Laws of Logarithms
- Solving Exponential and Logarithmic Equations
- Solving Equations Using Logs
- Logarithmic Equations and Expressions
- Past KCSE Questions on the Topic
Introduction
If y = ax then we introduce the inverse function logarithm and define loga y=x (Read as log base a of y equals x).
In general
y = ax ⇔ loga y=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 + log6 8 − log6 2
= log6 72 − log6 2
= log6 (72/2) = log6 36
= 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 = logb x
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 = n b. x = bm and y = bn c. xy = bm × bn d. xy = bm + n e. logb xy = m + n f. logb xy = logb x + logb y |
a. Setup b. Rewrite in exponent form c. Multiply together d. Product rule for exponents e. Rewrite in log form f. Substitution |
| 3. (quotient rule) logb x/y = logb x − logb y |
a. Let logb x = m and logb y = n b. x = bm and y = bn c. x/y = bm bn d. x = bm−n y e. logb x/y = m − n f. logb x/y = logb x − logb y |
a. Given: compact form b. Rewrite in exponent form c. Divide d. Quotient rule for exponents e. Rewrite in log form f. Substitution |
| 4. (power rule) logb xn = nlogb x |
a. Let m = logb x so x = bm b. xn = bmn c. logb xn = mn d. logb xn = nlogb x |
a. Setup b. Raise both sides to the nth power c. Rewrite as log d. Substitute |
| 5. Properties used to solve log equations: a. if bx = by, then x = y b. if logb x = logb y, then x= y |
a. This follows directly from the properties for exponents. b. i. logb x - logb y = 0 ii. logb x/y = 0 iii. x/y=b0 iv. x/y =1 so x = y |
b. i. Subtract from both sides ii. Quotient rule iii. Rewrite in exponent form iv. 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:
log5 (x+1) + log5 (x − 3) = 1 → (the 1 is the same as log5 5)
log5 (x+1)(x − 3) = log5 5
(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 5 + log5 1 = 1
1+0=1
log5 (-2+1) + log5 (-2 − 3) = 1
log5 -1 + log5 -5 not possible
Solving Equations Using Logs
Examples
- Solve the equation 10x =3.79
The definition of logs says if If y = ax then loga y=x or y = ax ⇔ loga y=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 - 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. - 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 - 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 4to give log (35 ÷ 46)
log 4 + 2log 3 log (4 × 32)
It does not really simplify things here but, in some cases, it can. - 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 ≈ 7/9 (taking 31.7 ≈ 32 = 9)
RHS = 2(52x+1) = 2(52×−0.686371+1) ≈ 2 × 5−0.4 ≈ 2/50.4 = 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
logx 8 = 3
x3 = 8
x3 =23
x = 2
Example
Solve log6 2
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= 20 thus x = 0
Example
Solve for x in (log10x)2 = 3 − log10x2
Solution
Let log10x = t……………………….(1)
Therefore t2 = 3 − 2t
t2 + 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
101 = x or 10-3= x
x = 10 or 1/1000
Note;
logb a = 1
loga b
Past KCSE Questions on the Topic
- Solve for (log3x)2 – ½ log3x = 3/2
- Find the values of x which satisfy the equation 52x – 6 (5x) + 5 =0
- Solve the equation
Log (x + 24) – 2 log 3 = log (9 – 2x) - Find the value of x in the following equation 49(x+1) + 7(2x) = 350
- Find x if 3 log 5 + log x2 = log 1/125
- Without using logarithm tables, find the value of x in the equation
Log x3 + log 5x = 5 log2 – log 2/5 -
- Given that P = 3y express the questions 3(2y -1) + 2 x 3(y-1) = 1 in terms of P
- Hence or otherwise find the value of y in the equation: 3(2y -1) + 2 x 3(y-1) = 1
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