Logarithms Class 9 Notes Advanced Maths Chapter 2

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Class 9 Maths Advanced Chapter 2 Logarithms Notes

Class 9 Advanced Maths Chapter 2 Notes

We already know that 3² = 9 and 3³ = 27, where 3 is the base and 2, 3 are the exponents. But, what if we know the base and the result, and want to find the exponent? This reverse process is called a logarithm.

For any positive number b (where, b > 0 and b ≠ 1) and a positive number a, if b raised to the power x equals a then the logarithm of a to the base b is x. We write this as
b^x = a ⇔ log_b a = x

• Base (b) The number which is raised to a power. It must be positive and not equal to 1.
• Argument (a) The number inside the logarithm. It must be positive.
  e.g. Consider 5³ = 125, here base = 5, exponent = 3 and result = 125. It can be written as log₅ 125 = 3.
  So, the two forms are equivalents 5³ = 125 (exponential) ⇔ log₅ 125 = 3 (logarithmic).

Example 1.
Write each of the following in the logarithmic form.
(i) 7³ = 343
(ii) 2^-10 = \(\frac{1}{1024}\)
(iii) 4^3/2 = 8
(iv) (0.2)³ = 0.008
Solution:
(i) We have, 7³ = 343
⇒ 3 = log₇343 [∵ a^m = n ⇒ m = log_a n]

(ii) We have, 2^-10 = \(\frac{1}{1024}\)
⇒ -10 = log₂ \(\frac{1}{1024}\) [∵ a^m = n ⇒ m = log_a n]

(iii) We have, 4^3/2 = 8
⇒ \(\frac{3}{2}\) = log₄8 [∵ a^m = n ⇒ m = log_a n]

(iv) We have, (0.2)³ = 0.008
⇒ 3 = log_0.2 (0.008) [∵ a^m = n ⇒ m = log_a n]

Example 2.
Find the values of x in each of the following.
(i) log₃ x = 3
(ii) log₄ x = \(\frac{3}{2}\)
(iii) log₁₀ x = – 3
(iv) log_x 16 = 2
(v) log₉ 27 = 2x + 3
Solution:
(i) We have, log₃ x = 3
⇒ x = (3)³ = 27 [∵ log_a n = m ⇒ a^m = n]

(ii) We have, log₄ x = \(\frac{3}{2}\)
⇒ \(x=(4)^{\frac{3}{2}}\) = 8 [∵ log_a n = m ⇒ a^m = n]

(iii) We have, log₁₀ x = – 3
⇒ x = (10)^-3 [∵ log_a n = m ⇒ a^m = n]
∴ x = \(\frac{1}{10^3}\) = \(\frac{1}{1000}\) [∵ a^-m = \(\frac{1}{a^m}\)]

(iv) We have, log_x 16 = 2
⇒ x² = 16 [∵ log_a n = m ⇒ a^m = n]
x = 4

(v) We have, log₉ 27 = 2x + 3
⇒ (9)²ⁿ⁺³ + 3 = 27 [∵ log_a n = m ⇒ a^m = n]
⇒ 3^4x+6 = 3³
⇒ 4x + 6 = 3
∴ x = –\(\frac{3}{4}\)

Example 3.
Find the logarithm in each of the following.
(i) 64 to the base 4
(ii) \(\frac{1}{81}\) to the base 27
(iii) 0.008 to the base 5
Solution:
(i) Let log₄ 64 = x then 64 = 4^x [∵ log_a n = m ⇒ a^m = n]
⇒ 4^x = 4 × 4 × 4
⇒ 4^x = 4³
On comparing the exponents of 4 both sides, we get
x = 3
Hence, log₄ 64 = 3.

(ii) Let log₂₇ \(\frac{1}{81}\) = x then \(\frac{1}{81}\) = (27)^x [∵ log_a n = m ⇒ a^m = n]
⇒ \(\frac{1}{3 \times 3 \times 3 \times 3}\) = (3 × 3 × 3)^x
⇒ (\(\frac{1}{3}\))⁴ = (3³)^x
⇒ 3^-4 = 3^3x [∵ \(\frac{1}{a^n}[latex] = a^-n and (a^m)ⁿ = a^mn ]
On comparing the exponents of 3 both sides, we get
3x = – 4
⇒ x = –[latex]\frac{4}{3}\)
Hence, log₂₇\(\frac{1}{81}\) = –\(\frac{4}{3}\).

(iii) Let log₅ 0.008 = x then
0.008 = 5^x [∵ log_a n = m ⇒ a^m = n]
⇒ \(\frac{8}{1000}\) = 5^x
⇒ \(\frac{1}{125}\) = 5^x
⇒ (5)^-3 = 5^x [∵ \(\frac{1}{a^n}\) = a^-n]
On comparing the exponents of 5 both sides, we get
x = -3
Hence, log₅ 0.008 = -3.

Laws of Logarithms and Their Uses

• Product Law log_a (m.n) = log_a m + log_a n
• Quotient Law log_a (\(\frac{m}{n}\)) = log_a m – log_a n
• Power Law log_a (m)ⁿ = n log_a m
• log_n m = \(\frac{\log _a m}{\log _a n}\) m > 0, n, a > 0, a ≠ 1, n ≠ 1
• a^log_am = m
• x^log_ay = y^log_ax
• log_a (1) = 0
• log_a (a) = 1
• Reciprocal Law log_a b = \(\frac{1}{\log _b a}\)
• log_aⁿ m^k = \(\frac{k}{n}\)log_a m

For positive values of m, n, x and a > 0, a ≠ 1
log_a (m + n) ≠ log_a m + log_a n
log_a (m – n) ≠ log_a m – log_a n
If x ≠ y then log_a n x ≠ log_a y

Common Logarithms (Base 10)

Logarithms in base 10 are called common logarithms. They are used in everyday scales like the richter scale (earthquakes), pH scale (acidity) and decibel scale (sound). We write log10 x simply as log x, assuming base 10.

Rules of Common Logarithms

Rule	Formula
Product	log(xy) = log x + log y
Quotient	log(\(\frac{x}{y}\)) = log x – log y
Power	log(xn) = n × log x
Special	log 1 = 0, log 10 = 1

Example 4.
Write it as a single logarithm.
(a) log₅ 4 + log₅ 6
(b) log₃ 16 – log₃ 2
(c) log₆ 4 + log₆ 9 – 2log₆ 3
(d) 1 + log₄ 7
Solution:
We have, log₅ 4 + log₅ 6
= log₅(4 × 6) [∵ log_a m + log_a n = log_a mn]
= log₅ 24

(b) We have, log₃16 – log₃ 2
= log₃(\(\frac{16}{2}\)) [∵ log_a m – log_a n = log_a \(\frac{m}{n}\)]
= log₃ 8

(c) We have, log₆ 4 + log₆ 9 – 2log₆ 3
= log₆ (4 × 9) – log₆ 3² [∵ log m + log n = log mn and nlog m = log mⁿ]
= log₆ 36 – log₆ 9
= log₆(\(\frac{36}{9}\)) [∵ log m – log n = log \(\frac{m}{n}\)]
= log₆ 4

(d) We have, 1 + log₄ 7
= log₄ 4 + log₄ 7 [∵ 1 = log_a a]
= log₄ (4 × 7) [∵ log m + log n = log mn]
= log₄ 28

Example 5.
Find the value of
(a) log₆ 216
(b) log₆ 32√2
Solution:
(a) We have, log₆ 216
= log₆ 6³
= 3log₆ 6 [∵ log_a mⁿ = nlog_a m]
= 3(1) [∵ log_a a = 1]
= 3

(b) We have, log₆ 32√2

Example 6.
Simplify \(\frac{\log _3 8}{\log _9 16 \log _4 10}\)
Solution:

Example 7.
Find the value of
log₃2.log₄3.log₅4. ….. log₈7.
Solution:
We have,
log₃2.log₄3.log₅4. ….. log₈7.
By base changing formula,
log₃2.log₄3.log₅4. ….. log₈7 = \(\frac{\log 2}{\log 3} \cdot \frac{\log 3}{\log 4} \cdot \frac{\log 4}{\log 5}\) … \(\frac{\log 2}{\log 3} \cdot \frac{\log 3}{\log 4} \cdot \frac{\log 4}{\log 5} \ldots \cdot \frac{\log 7}{\log 8}=\frac{\log 2}{\log 8}=\frac{\log 2}{3 \log 2}=\frac{1}{3}\)

Example 8.
If log₁₀ 2 = m and log ₁₀3 = n then express each of the following in terms of m and n.
(i) log₁₀ 90
(ii) log₁₀ 0.54
Solution:
Given, log₁₀ 2 = m and log₁₀ 3 = n

(i) We have, log₁₀90 = log₁₀ (10 × 9)
= log₁₀10 + log₁₀ 3² [∵ log mn = log m + log n ]
= 1 + 21og₁₀ 3 [∵ log mⁿ = nlog m]
= 2 log₁₀ 3 + 1 = 1 + 2n [∵ log_a a = 1]

(ii) log₁₀ 0.54 = log₁₀\(\frac{54}{100}\) = log₁₀ \(\frac{27 \times 2}{100}\)
= log₁₀ 27 × 2 – log₁₀ 100 [∵ log \(\frac{m}{n}\) = log m – log n]
= log₁₀ 27 + log₁₀ 2 – log₁₀ 100 [∵ log mn = log m + log n]
= log₁₀ 3³ + log₁₀2 – log₁₀10²
= 3log₁₀ 3 + log₁₀ 2 – 2log₁₀ 10 [∵ log mⁿ = nlog m]
= 3n + m – 2 [∵ log₁₀ 10 = 1]
= m + 3n – 2

Example 9.
Determine the logarithmic expansion of y = \(\frac{5^p 7^q}{3^r}\)
Solution:
Given, y = \(\frac{5^p 7^q}{3^r}\)
On taking log both sides, we get
log y = log\(\frac{5^p 7^q}{3^r}\)
⇒ log y = log5^p7^q – log 3^r [∵ log \(\frac{m}{n}\) = log m – log n ]
log y = log5^p + log7^q – log3^r [∵ log mn = log m + log n]
log y = p log5 + q log7 – r log3 [∵ nlog m = logmⁿ]
Hence, log y = p log5 + q log7 – r log3 is the logarithmic
expansion of y = \(\frac{5^p 7^q}{3^r}\).

Example 10.
Prove that log_3⁴ 4³ = \(\frac{3}{4}\)log₃ 4.
Solution:
Let log_3⁴ 4³ = x
⇒ (3⁴)^x = 4³ [∵ a^m = n ⇒ log_a n = m]
⇒ 3^4x = 4³
On taking log with base 3 both sides, we get
⇒ log₃ 3^4x = log₃ 4³
⇒ 4x = log₃ 4³
⇒ 4x = 31og₃ 4 ⇒ x = \(\frac{3}{4}\)log₃ 4
∴ LHS = RHS

Example 11.
Solve the equations for x
log_x 2 × log_x/16 2 = log_x/64 2
Solution:
Given, log_x 2 × log_x/16 2 = log_x/64 2
⇒ \(\frac{1}{\log _2 x}\) × \(\frac{1}{\log _2\left(\frac{x}{16}\right)}\) = \(\frac{1}{\log _2\left(\frac{x}{64}\right)}\) [∵ log_a b = \(\frac{1}{\log _b a}\)]
By cross multiplication, we get
⇒ log₂(x/64) = log₂ x × log₂(x/16)
⇒ log₂ x – log₂ 64 = log₂ x(log₂ x – log₂ 16) [∵ log_a\(\frac{m}{n}\) = log_a m – log_an]
⇒ log₂ x – log₂ 2⁶ = log₂ x(log₂ x – log₂2⁴)
⇒ log₂ x – 6 log₂ 2 = log₂ x(log₂ x – 4 log₂ 2)
[∵ log_a mⁿ = nlog_a m]
⇒ log₂ x – 6 × 1 = log₂ x(log₂ x – 4 × 1) [∵ log_a a = 1]
⇒ (log₂ x)² – 5log₂ x + 6 = 0
Let y = log₂ x
⇒ y² – 5y + 6 = 0 ……. (i)
⇒ y² – 3y – 2y + 6 = 0
= y(y – 3) – 2(y – 3) = 0
⇒ (y – 3)(y – 2) = 0 ⇒ y = 3 and y = 2
On putting the values ofy in Eq. (i), we get
log₂ x = 3 and log₂ x = 2
⇒ x = 2³ and x = 2²
⇒ x = 8
or x = 4.
Hence, the solutions of the given equation are 8 and 4.

Example 12.
Simplify the following.
(i) log 4 + log 25
(ii) log 1000 – log 10
(iii) log (10³)
Solution:
(i) We have, log 4 + log 25
= log(4 × 25) = log 100 [∵ log m + log n = log mn]
= log 10² = 2 log 10 [∵ log mⁿ = nlog m]
= 2(1) = 2 [∵ log 10 = 1]

(ii) log 1000 – log 10
= log(\(\frac{1000}{10}\)) = log 100 [∵ log m – log n = log \(\frac{m}{n}\)]
= log 10² = 2 log 10 [log mⁿ = nlog m]
= 2(1) = 2 [∵ log 10 = 1]
(iii) log (10)³ = 31og 10 [∵ log mⁿ = nlog m]
= 3(1) = 3 [∵ log 10 = 1]

Example 13.
Evaluate the following.
(i) \(\frac{\log 8 \log 9}{\log 27}\)
(ii) \(\frac{\log 625}{\log 125}\)
(iii) 3 + \(\frac{1}{2}\)log₁₀9 – 2log₁₀5
(iv) 4 + log₁₀(0.001)
Solution:

(iii) We have, 3 + \(\frac{1}{2}\) log₁₀ 9 – 2log₁₀ 5
= 3 × 1 + \(\frac{1}{2}\)log (3 × 3) – 2log₁₀ 5
= 3log₁₀ 10 + \(\frac{1}{2}\) log₁₀ 3² – 2log₁₀ 5 [∵ 1 = log₁₀ 10]
= log₁₀ 10³ + \(\log _{10} 3^{2 \times \frac{1}{2}}\) – log₁₀ (5)² [∵ nlog m = log mⁿ]
= log₁₀ 1000 + log₁₀ 3 – log₁₀ 25
= log₁₀ (1000 × 3) – log₁₀ 25 [∵ log m + log n = log mn]
= log₁₀ \(\frac{3000}{25}\) [∵ log m – log n = log \(\frac{m}{n}\)]
= log₁₀ 120

(iv) 4 + log₁₀ (0.001)
= 4 + log₁₀(10)^-3
= 4 – 3log₁₀ 10 [by power law]
= 4 – 3 × 1 [∵ log₁₀ 10 = 1]
= 1

Example 14.
If log₁₀ 8 = 0.90 then find the value of log \(\sqrt{32},\).
Solution:
We have, log₁₀ 8 = 0.90
⇒ log₁₀ 2³ = 0.90 ⇒ 3log₁₀ 2 = 0.90
⇒ log₁₀ 2 = \(\frac{0.90}{3}\) = 0.30 …………(i)
Now, log = \(\sqrt{32},\) log₁₀(32)^1/2 = \(\frac{1}{2}\)log₁₀ 32 [∵ log mⁿ = nlog m]
= \(\frac{1}{2}\)log₁₀ 2⁵ = \(\frac{5}{2}\) log₁₀ 2
= \(\frac{5}{2}\) × 0.30 = 5 × 0.15 = 0.75 [from Eq. (i)]

Example 15.
Express log4.5 + log\(\frac{25}{50}\) in terms of log 2 and log 3.
Solution:
We have, log4.5 + log\(\frac{25}{50}\)
= log\(\frac{45}{10}\) + log\(\frac{25}{50}\) = log\(\frac{9}{2}\) + log\(\frac{1}{2}\)
= log9 – log2 + log1 – log2 [∵ log\(\frac{m}{n}\) = logm – logn
= log3² – 2log2 [∵ log1 = 0]
= 2log2 – 2log2 [∵ logmⁿ = nlog m]
= 2(log3 – log2)

Example 16.
If log 7 – log2 + log 16 – 2log 3 – log\(\frac{7}{45}\) = 1 + log x then find x.
Solution:
We have,
log7 – log2 + logl6 -2log3 – log\(\frac{7}{45}\) = 1 + logx
⇒ log7 – log2 + log16 – log3² – log\(\frac{7}{45}\) = 1 + logx [∵ n log m = log mⁿ]
⇒ log7 – log2 + log16 – log9 – (log7 – log45) = 1 + logx [∵ log\(\frac{m}{n}\) = logm – logn].
⇒ log7 – log2 + log 16 – log9 – log7 + log45 = 1 + logx
⇒ (log16 + log45) – (log2 + log9) = 1 + logx
⇒ log(16 × 45) – log(2 × 9) = 1 + logx [∵ logm + logn = logmn]
⇒ log \(\frac{16 \times 45}{2 \times 9}\) = 1 + logx [∵ logm – logn = log\(\frac{m}{n}\)]
⇒ log40 = 1 + logx = log10 + logx [∵ 1 = log10]
⇒ log40 = log10x
On comparing the coefficient of log both sides, we get
10x = 40 ⇒ x = 4

Example 17.
Solve for x, if log₁₀(x² – 21) = 2.
Solution:
We have, log₁₀(x² – 21) =2
⇒ x² – 21 = (10)² [∵ log_a n = m ⇒ a^m = n]
⇒ x² – 21 = 100 ⇒ x² = 100 + 21 = 121
∴ x² = (11)² ⇒ x = ± 11 [taking square root]

Example 18.
Solve for x, if log (x + 3) – log (x – 3) = 1.
Solution:
We have, log(x + 3) – log(x – 3) = 1
⇒ log(\(\frac{x+3}{x-3}\)) = 1 [∵ log m – log n = log \(\frac{m}{n}\)]
⇒ log(\(\frac{x+3}{x-3}\)) = log10 [∵ 1 = log10]
⇒ (\(\frac{x+3}{x-3}\)) = 10
⇒ x + 3 = 10(x – 3)
⇒ x + 3= 10x – 30
⇒ x – 10x = – 30 – 3
⇒ -9x = -33
∴ x = \(\frac{33}{9}\) = \(\frac{11}{3}\)
= 3\(\frac{2}{3}\)

Example 19.
Solve for x, if (log₁₀ x)² – (log₁₀ x) – 2 = 0.
Solution:
We have, (log₁₀ x)² – (log₁₀ x) – 2 = 0
Let log₁₀ x = t then the given expression reduces to
t² – t – 2 = 0.
Now, factorise by splitting the middle term, we get
t² – 2t + t – 2 = 0
⇒ t(t – 2) + 1(t – 2) = 0
⇒ (t – 2) (t + 1) = 0
t – 2 = 0 or t + 1 = 0
⇒ t = 2 or t = – 1
On putting the value of t, we get
log₁₀ x = -1 or log₁₀ x = 2
⇒ x = (10)^-1 or x = (10)² [∵ log_a n = m ⇒ a^m = n]
Hence, x = \(\frac{11}{3}\) and 100.

Example 20.
If p = log_x yz, q = log_y zx and r = log_z xy
then prove that \(\frac{1}{1+p}+\frac{1}{1+q}+\frac{1}{1+r}\) = 1.
Solution:

= log_xyz x + log_xyz y + log_xyz z
= log_xyz xyz = 1 [∵ log m + log n = log mn]
= RHS
∴ LHS = RHS

Common logarithms can scale both very small and very large numbers into a manageable range, e.g. Numbers  ranging from 0.0000000001 to 10000000000 simply become H 10 to 10 in log form.

Natural Logarithm

The natural logarithm is the logarithm to the base e, where e is an irrational number (2 < e < 3). It is denoted by log_e x or ln x.

Applications of Logarithms in Daily Life

Logarithms are mathematical tools used to simplify calculations involving very large or very small quantities and help in understanding exponential changes.

In Chemistry, the pH scale uses logarithms to measure acidity, given by pH = -log₁₀[H⁺], making very small concentrations easier to handle.

In Music, when a note increases by one octave, its frequency doubles, yet we perceive it as equal steps, This shows that sound is heard logarithmically.

In Social Science, population growth follows the form P = P₀(1 + r)^t (and logarithms are used to calculate the time required for population to double or reach a certain size.

In Geography, the. richter scale is logarithmic, where an increase of one unit represents a tenfold increase in amplitude.

Example 21.
The number of bacteria N in a culture after t h is given by t = 5 × log 2(\(\frac{N}{100}\)).
How many hours will it take for the bacteria population to reach 1600?
Solution:
We have, t = 5 × log₂(\(\frac{N}{100}\))
⇒ t = 5 × loglog2(\(\frac{1600}{100}\)) ⇒ t = 5 × log₂ 16
⇒ t = 5 × log₂ (2⁴)
⇒ t = 5 × 4 log₂ 2 [∵ log_a m^k = klog_a m]
⇒ t = 20(1) [∵ log_a a = 1]
⇒ t = 20 h

Example 22.
A solution has pH 2 and another has pH 5. How many times more acidic is the first solution?
[PH = -log₁₀[H⁺]]
Solution:
We have, pH= -log₁₀[H⁺]]
For the first solution,
2 = \(-\log _{10}\left[\mathrm{H}_1^{+}\right]=\log _{10}\left[\mathrm{H}_1^{+}\right]\) = -2
= \(\left[\mathrm{H}_1^{+}\right]\) = 10^-2 ……. (i)
For the second solution,
5 = \(-\log _{10}\left[\mathrm{H}_2^{+}\right]=\log _{10}\left[\mathrm{H}_2^{+}\right]\) = -5
= \(\left[\mathrm{H}_2^{+}\right]\) = 10^-5 ……. (ii)
To find how many times more acidic, we take the ratio
\(\frac{\left[\mathrm{H}_1^{+}\right]}{\left[\mathrm{H}_2^{+}\right]}\) = \(\frac{10^{-2}}{10^{-5}}\) = 10^-2-(-5) = 10³ = 1000
Hence, the first solution is 1000 times more acidic.

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