Exercise 1.5 — Logarithms

Logarithms, their properties and related problems.

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Exercise 1.5 is the final exercise of Chapter 1, Real Numbers, in Class 10 Mathematics (CBSE, Telangana & Andhra Pradesh syllabus). It introduces one of the most useful tools in the whole of mathematics — the logarithm — and asks you to apply six fundamental properties of logarithms to evaluate expressions, expand them, rewrite combinations of logs as a single logarithm, and solve equations involving exponents.

Logarithms might look unfamiliar at first, but they are simply a different way of writing the same statement you already know from exponents and powers. Below you'll find the complete definition, every property with its proof, and a fully worked, step-by-step explanation of all the problems in this exercise — ideal for board exam revision.

Definition of a Logarithm 6 Core Properties (with Proofs) Common vs Natural Logarithms 29 Solved Problems
💡 Core idea of this exercise: A logarithm answers one simple question — "to what power must the base be raised to get this number?" Once you can move comfortably between the exponential form aˣ = N and the logarithmic form x = logₐ N, every problem in this exercise becomes a matter of applying one of just six properties.

What Is a Logarithm?

If a is a positive number (other than 1) and aˣ = N, then we say that x is the logarithm of N to the base a, and we write this as x = loga N. The two forms — exponential and logarithmic — describe exactly the same relationship between three numbers; only the "subject" of the statement changes.

aˣ = N Exponential Form x = logₐN Logarithmic Form same fact, two forms

Every exponential statement aˣ = N can be rewritten instantly as a logarithmic statement x = logₐN, and vice versa.

If aˣ = N, then x = logₐ N — where a > 0, a ≠ 1, N > 0, and a, N are real numbers.

For example, since 2³ = 8, we can immediately say log₂ 8 = 3. Since 10² = 100, we know log₁₀ 100 = 2. Getting comfortable with this back-and-forth conversion is the single most important skill for this exercise.

Common Logarithms and Natural Logarithms

Logarithms can technically be taken to any valid base, but two particular bases are used so often that they have their own names and special notation:

Common Logarithm

  • Base = 10
  • Written simply as log N (the base 10 is left out)
  • Example: log 7, log 16
  • Used in scientific scales like pH and decibels

Natural Logarithm

  • Base = e (≈ 2.718)
  • Written as logₑ N or ln N
  • Example: logₑ 5, logₑ 19
  • Used heavily in higher mathematics, calculus, and science

In this exercise, whenever you see "log N" with no base written, you should always read it as log₁₀ N — the common logarithm.

  • pH scale (chemistry): pH is defined using a common logarithm of the hydrogen-ion concentration — each whole number change in pH represents a tenfold change in acidity.
  • Richter scale (earthquakes): Earthquake magnitude is measured on a logarithmic scale, so a magnitude 6 earthquake releases far more than twice the energy of a magnitude 3 one.
  • Decibel scale (sound): Sound intensity is also measured logarithmically, which is why a "small" increase in decibels can represent a large real increase in loudness.
📌 Why use logarithms at all? Quantities in nature — earthquake energy, sound intensity, acidity — often vary across an enormous range of values. Logarithms compress these huge ranges into small, manageable numbers, which is exactly why scientists reach for them.

The 6 Fundamental Properties of Logarithms

Every problem in this exercise is solved using one (or a combination) of the following six properties. Each one follows directly from the laws of exponents you already know — here is the full proof of each.

Property 1 — Product Rule
logₐ (mn) = logₐ m + logₐ n
Let logₐ m = x and logₐ n = y ⟹ m = aˣ, n = aʸ mn = aˣ · aʸ = aˣ⁺ʸ ⟹ logₐ (mn) = x + y = logₐ m + logₐ n
Property 2 — Quotient Rule
logₐ (m/n) = logₐ m − logₐ n
Let logₐ m = x and logₐ n = y ⟹ m = aˣ, n = aʸ m/n = aˣ/aʸ = aˣ⁻ʸ ⟹ logₐ (m/n) = x − y = logₐ m − logₐ n
Property 3 — Power Rule
logₐ (mⁿ) = n · logₐ m
Let m = aˣ ⟹ mⁿ = (aˣ)ⁿ = aⁿˣ ⟹ logₐ (mⁿ) = nx = n · logₐ m (since x = logₐ m)
This is the most-used property in the whole exercise — it lets you pull an exponent out in front of a logarithm.
Properties 4, 5 & 6 — The Basic Identities
  • logₐ a = 1 — because a¹ = a, so by definition log of a to its own base is 1.
  • logₐ 1 = 0 — because a⁰ = 1 for any valid base a, so the log of 1 (in any base) is always 0.
  • alogₐ m = m — if logₐ m = x, then by definition aˣ = m, so substituting back gives alogₐ m = m directly.
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Solved Examples — Question 1: Determine the Value

In each part below, the strategy is the same: rewrite the number inside the logarithm as a power of the base (using roots, negative exponents, or factorisation), then apply Property 3 followed by Property 4.

(i)
log₂₅ 5
5² = 25 ⟹ 5 = 25^(1/2)
log₂₅25^(1/2) = (1/2)·log₂₅25
Answer = 1/2
(ii)
log₈₁ 3
3⁴ = 81 ⟹ 3 = 81^(1/4)
log₈₁81^(1/4) = (1/4)·log₈₁81
Answer = 1/4
(iii)
log₂ (1/16)
1/16 = 16⁻¹ = (2⁴)⁻¹ = 2⁻⁴
log₂2⁻⁴ = −4·log₂2
Answer = −4
(iv)
log₇ 1
Directly by Property 5: logₐ 1 = 0
Answer = 0
(v)
logx √x
√x = x^(1/2)
logxx^(1/2) = (1/2)·logxx
Answer = 1/2
(vi)
log₂ 512
512 = 2×2×2×2×2×2×2×2×2 = 2⁹
log₂2⁹ = 9·log₂2
Answer = 9
(vii)
log₁₀ 0.01
0.01 = 1/100 = (10²)⁻¹ = 10⁻²
log₁₀10⁻² = −2·log₁₀10
Answer = −2
(viii)
log3/2 (8/27)
8/27 = 2³/3³ = (2/3)³ = (3/2)⁻³
log3/2(3/2)⁻³ = −3·log3/2(3/2)
Answer = −3
Spotlight — Part (ix)
22 + log₂3

This one mixes ordinary exponent rules with logarithm Property 6, so it deserves a closer look.

22+log₂3 = 2² · 2log₂3 [∵ aᵏ⁺ˡ = aᵏ·aˡ] = 4 × 3 [∵ alogₐ m = m, Property 6] = 12

Solved Examples — Question 2: Write as a Single Logarithm

These problems run Properties 1, 2 and 3 in reverse — combining several logarithms into a single "log N" expression, then evaluating it where possible.

(i)
log 2 + log 5
= log (2×5) = log 10
= log 10 = 1
(ii)
log₂16 − log₂2
= log₂(16/2) = log₂8 = log₂2³
= 3
(iii)
3 log₆₄ 4
= log₆₄4³ = log₆₄64
= 1
(iv)
2 log 3 − 3 log 2
= log 3² − log 2³ = log 9 − log 8
= log (9/8)
(v)
log 10 + 2 log 3 − log 2
= log 10 + log 9 − log 2 = log (10×9/2)
= log 45

Solved Examples — Question 3: Evaluate in Terms of x and y

Here you're given x = log₂3 and y = log₂5, and asked to express other base-2 logarithms in terms of x and y. The trick is to factorise the number inside the log into 2's, 3's and 5's, then convert the product into a sum using Property 1.

60 2 30 2 15 3 5

Factor tree for 60: 60 = 2 × 2 × 3 × 5 = 2² × 3 × 5 — used in part (iii) below.

(i)
log₂ 15
15 = 3 × 5
log₂15 = log₂3 + log₂5
= x + y
(ii)
log₂ 7.5
7.5 = 15/2
log₂15 − log₂2 = (x+y) − 1
= x + y − 1
(iii)
log₂ 60
60 = 2² × 3 × 5
= 2log₂2 + log₂3 + log₂5
= 2 + x + y
(iv)
log₂ 6750
6750 = 2 × 3³ × 5³
= log₂2 + 3log₂3 + 3log₂5
= 1 + 3x + 3y

Solved Examples — Question 4: Expand the Logarithm

"Expanding" means going the other way — breaking one complicated logarithm into a sum or difference of several simpler ones, using Properties 1, 2 and 3 together.

(i)
log 1000
1000 = 2³ × 5³
= log 2³ + log 5³
= 3 log 2 + 3 log 5
(ii)
log (128/625)
128 = 2⁷, 625 = 5⁴
= log 2⁷ − log 5⁴
= 7 log 2 − 4 log 5
(iii)
log (x²y³z⁴)
= log x² + log y³ + log z⁴
= 2 log x + 3 log y + 4 log z
(iv)
log (p²q³/r⁴)
= log p² + log q³ − log r⁴
= 2 log p + 3 log q − 4 log r
(v)
log √(x³/y²)
= (1/2)[log x³ − log y²]
= (1/2)(3 log x − 2 log y)
= (3/2) log x − log y

Solved Example — Question 5: A Proof Using Logarithm Properties

Question

If x² + y² = 25xy, prove that 2 log (x + y) = 3 log 3 + log x + log y.

Given: x² + y² = 25xy Add 2xy to both sides: x² + y² + 2xy = 25xy + 2xy = 27xy ⟹ (x + y)² = 27xy Taking log of both sides: log (x+y)² = log (27xy) 2 log (x+y) = log 27 + log x + log y [Properties 3 & 1] = log 3³ + log x + log y ⟹ 2 log (x+y) = 3 log 3 + log x + log y — proved.

Solved Example — Question 6: Finding the Value of x/y + y/x

Question

If log[(x+y)/3] = (1/2)(log x + log y), find the value of x/y + y/x.

2 log [(x+y)/3] = log x + log y = log (xy) [Property 1] ⟹ log [(x+y)/3]² = log (xy) [Property 3] ⟹ (x+y)²/9 = xy ⟹ (x+y)² = 9xy ⟹ x² + 2xy + y² = 9xy ⟹ x² + y² = 7xy Dividing both sides by xy: x/y + y/x = 7

Solved Example — Question 7: Logarithmic Equation With Two Unknowns

Question

If 2.3ˣ = 0.23ʸ = 1000, find the value of 1/x − 1/y.

2.3ˣ = 1000 ⟹ 1/x = log₁₀₀₀ 2.3 0.23ʸ = 1000 ⟹ 1/y = log₁₀₀₀ 0.23 1/x − 1/y = log₁₀₀₀ 2.3 − log₁₀₀₀ 0.23 = log₁₀₀₀ (2.3/0.23) = log₁₀₀₀ 10 10 = 1000^(1/3) ⟹ log₁₀₀₀ 1000^(1/3) = 1/3 ⟹ 1/x − 1/y = 1/3

Solved Example — Question 8: Solving an Exponential Equation Using Logs

Question

If 2ˣ⁺¹ = 3¹⁻ˣ, find the value of x.

Taking log of both sides: (x+1) log 2 = (1−x) log 3 x log 2 + log 2 = log 3 − x log 3 x log 2 + x log 3 = log 3 − log 2 x (log 2 + log 3) = log 3 − log 2 ⟹ x = (log 3 − log 2) / (log 3 + log 2)
📌 Taking logs is the standard technique whenever the variable appears as an exponent on both sides of an equation, and the bases are different.

Solved Examples — Question 9: Rational or Irrational?

Part (i)

Is log 2 rational or irrational? Justify your answer.

Assume log 2 is rational: log 2 = p/q, where p, q are integers, q ≠ 0. Since log 1 = 0 and log 10 = 1, and 0 < log 2 < 1, we get 0 < p/q < 1, so 0 < p < q. log 2 = p/q ⟹ 2 = 10^(p/q) ⟹ 2^q = 10^p = 2^p · 5^p ⟹ 2^(q−p) = 5^p Since q > p, the left side 2^(q−p) is always even. But 5^p (with p ≥ 1) is always odd. An even number can never equal an odd number — contradiction! ⟹ log 2 is irrational.
Part (ii)

Is log 100 rational or irrational? Justify your answer.

log 100 = log₁₀ 10² = 2 · log₁₀10 [Property 3] = 2 × 1 = 2 Since 2 can be written as 2/1, log 100 is a rational number.

Common Mistakes to Avoid

  • Confusing logₐ(m + n) with logₐ m + logₐ n: There is no property for the logarithm of a sum — the product rule only applies to multiplication, never to addition.
  • Forgetting the base when it isn't written: "log N" with no subscript always means base 10 — don't confuse it with "ln N" (base e).
  • Mishandling negative exponents: Remember that 1/aⁿ = a⁻ⁿ before applying Property 3 — skipping this step is the most common error in problems like Q1 (iii) and (vii).
  • Dropping the conditions on the base: A logarithm is only defined when the base a > 0, a ≠ 1, and the argument N > 0 — always check these conditions hold in equation-solving problems.
  • Sign errors while expanding: When expanding a quotient inside a log, every term in the denominator gets a minus sign — easy to miss with three or more terms, as in Q4 (iv).
Exam tip: Always state which property you are using at each step (just like in the solutions above) — in CBSE, Telangana and AP board exams, marks are awarded for correctly identifying and applying each logarithm property, not just for the final numeric answer.

Quick Reference — All 6 Properties at a Glance

#PropertyOne-Line Reason
1logₐ(mn) = logₐm + logₐnMultiplying numbers adds their exponents
2logₐ(m/n) = logₐm − logₐnDividing numbers subtracts their exponents
3logₐ(mⁿ) = n·logₐmRaising to a power multiplies the exponent
4logₐ a = 1a¹ = a, by definition
5logₐ 1 = 0a⁰ = 1 for any base a
6a^(logₐ m) = mThe log and the exponential "undo" each other

What This Lesson Prepares You For

Logarithms wrap up the Real Numbers chapter by tying together everything you've learned about exponents, powers, and the rational/irrational classification of numbers — revisiting the introduction to real numbers alongside this exercise can help connect the two ideas, especially for the rational/irrational proofs in Question 9.

The algebraic manipulation skills you've practised here — combining, expanding, and solving equations with logs — also build the foundation you'll need later for exponential and logarithmic functions, and they reinforce the factorisation techniques used throughout Polynomials and Quadratic Equations, the next two chapters in the Class 10 syllabus.

📐 Board Exam Tip (CBSE, Telangana & AP): Logarithm questions are a reliable, high-scoring exam topic because they follow a small, fixed set of rules. Practising every part of Q1–Q4 until you can do them without referring back to the properties table is the single best way to secure full marks on this exercise.
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