MathematicsClass 9ICSE

Turn Multiplication into Addition and Powers into Products

Derive the product, quotient and power laws from the laws of indices, expand a messy logarithm into separate terms, collapse a sum of logs into one, and prove identities and solve equations safely.

Why do logarithms turn multiplication into addition?

Because indices already do, and a logarithm is an index.

Watch it happen on numbers you know:

lo g_{2} 8 + lo g_{2} 4 = 3 + 2 = 5, and lo g_{2} (8 \times 4) = lo g_{2} 32 = 5

A multiplication on the inside became an addition on the outside. That is not a coincidence of these numbers — it is the addition law of indices,

a^{m} \times a^{n} = a^{m + n}

, seen from the other side. The indices were

3

and

2

; multiplying the powers added them; and the logarithms are those indices.

Every law in this chapter is one line of the indices chapter, re-read. That is worth knowing, because it means you can rebuild any of them if you forget it.

And it explains why the laws stop where they do. There is no rule for

lo g (M + N)

, because there is no rule for

a^{m} + a^{n}

. The gap in one chapter is the same gap in the other.

This page covers the second part of the ICSE Class 9 Mathematics chapter on logarithms: the product, quotient and power laws, expanding an expression into separate terms, collapsing a sum or difference into a single logarithm, and proving identities and solving equations.

Formula

What are the three laws of logarithms and where does each come from?

A product becomes a sum, a quotient becomes a difference, and an index becomes a multiplier.

lo g_{a} (M N) = lo g_{a} M + lo g_{a} N

lo g_{a} \frac{M}{N} = lo g_{a} M - lo g_{a} N

lo g_{a} M^{p} = p lo g_{a} M

All three need

M

and

N

positive and the base legal — positive and not

1

.

Where the product law comes from, in three lines. Let

lo g_{a} M = x

and

lo g_{a} N = y

. Converting to exponential form gives

M = a^{x}

and

N = a^{y}

, so

M N = a^{x} \times a^{y} = a^{x + y}

Converting back,

lo g_{a} (M N) = x + y = lo g_{a} M + lo g_{a} N

. The middle step is the addition law of indices and nothing else.

The quotient law follows the same way from

\frac{a ^{x}}{a ^{y}} = a^{x - y}

, and the power law from

(a^{x})^{p} = a^{x p}

.

Two values you use constantly, both straight from Part 1:

lo g_{a} a = 1 lo g_{a} 1 = 0

Worked example with numbers, to see all three at once. Take base

10

, and use only

lo g 2 \approx 0.301

and

lo g 3 \approx 0.477

:

-

lo g 6 = lo g 2 + lo g 3 \approx 0.778

lo g 1.5 = lo g 3 - lo g 2 \approx 0.176

lo g 8 = 3 lo g 2 \approx 0.903

lo g 5 = lo g \frac{10}{2} = 1 - lo g 2 \approx 0.699

Four different logarithms from two numbers. That compression is exactly what the laws are for, and the last line is the one worth remembering: **

lo g 5

is easier to get from

lo g 2

than from anything else.

The three false laws, which the examiner is hoping for.**

-

lo g (M + N)

is not

lo g M + lo g N

. Test:

lo g_{2} (4 + 4) = 3

but

lo g_{2} 4 + lo g_{2} 4 = 4

\frac{lo g M}{lo g N}

is not

lo g M - lo g N

. The quotient law needs the division inside the logarithm
-

(lo g M)^{2}

is not

2 lo g M

. The power law needs the index inside, on

M

How do you expand a logarithm of a messy fraction into separate terms?

Work outwards: split the division first, then the multiplications, then bring every index down in front.

Worked example 1. Expand

lo g \frac{a ^{4} b ^{2}}{c ^{3}}

.

Split the quotient:

lo g \frac{a ^{4} b ^{2}}{c ^{3}} = lo g (a^{4} b^{2}) - lo g c^{3}

Split the product, then apply the power law to each term:

= lo g a^{4} + lo g b^{2} - lo g c^{3} = 4 lo g a + 2 lo g b - 3 lo g c

Every index has turned into a coefficient, which is the whole point of expanding.

Worked example 2 — a root on the outside. Expand

lo g \frac{x ^{3} y}{z ^{5}}

.

A square root is the index

\frac{1}{2}

, so deal with that first and it multiplies everything that follows:

lo g (\frac{x ^{3} y}{z ^{5}})^{1/2} = \frac{1}{2} lo g \frac{x ^{3} y}{z ^{5}} = \frac{1}{2} (3 lo g x + lo g y - 5 lo g z)

The bracket is compulsory. Dropping it turns the

\frac{1}{2}

into a coefficient of the first term only, and the last two terms come out wrong by a factor of two.

Worked example 3 — numbers rather than letters. Express

lo g \frac{48}{25}

in terms of

lo g 2

lo g 3

and

lo g 5

.

Factorise both parts into primes before doing anything else:

48 = 2^{4} \times 3

and

25 = 5^{2}

. So

lo g \frac{48}{25} = 4 lo g 2 + lo g 3 - 2 lo g 5

Check numerically:

4 (0.301) + 0.477 - 2 (0.699) = 1.204 + 0.477 - 1.398 = 0.283

, and

lo g \frac{48}{25} = lo g 1.92 \approx 0.283

, as required.

The order that keeps signs safe is division before multiplication. If you bring indices down first and split afterwards, the minus sign in front of

lo g c^{3}

has to be distributed over several terms, and that is where a sign gets lost. Split the fraction while it is still a single object, and each term arrives with its sign already attached.

How do you collapse a sum of logarithms into one and evaluate it?

Push every coefficient back up as an index, then read plus as multiply and minus as divide. It is the previous section run backwards.

Worked example 1. Express

2 lo g 3 + 3 lo g 2 - lo g 6

as a single logarithm.

Coefficients go back up first:

= lo g 3^{2} + lo g 2^{3} - lo g 6 = lo g 9 + lo g 8 - lo g 6 = lo g \frac{9 \times 8}{6} = lo g 12

Worked example 2 — an answer that is a whole number. Evaluate

lo g 2 + lo g 5

in base

10

lo g 2 + lo g 5 = lo g 10 = 1

No tables needed. Any pair of logs whose product is a power of ten collapses to an integer, and

2 \times 5

is the pair the examiner uses most.

Worked example 3. Evaluate

\frac{1}{2} lo g 25 + lo g 4 - lo g 2

in base

10

\frac{1}{2} lo g 25 = lo g 2 5^{1/2} = lo g 5

lo g 5 + lo g 4 - lo g 2 = lo g \frac{5 \times 4}{2} = lo g 10 = 1

Worked example 4. Evaluate

lo g_{10} 25 + lo g_{10} 4

= lo g_{10} 100 = 2

Check by the definition:

1 0^{2} = 100

, as required.

Worked example 5 — a different base. Evaluate

lo g_{2} 48 - lo g_{2} 3

= lo g_{2} \frac{48}{3} = lo g_{2} 16 = 4

The tactic that turns these into one-liners is looking at the numbers before touching the laws. If the combination can be steered towards

10

100

or a power of the base, the answer is an integer and the tables are irrelevant. Aim for the base's own powers and the question usually collapses.

One boundary case.

lo g M + lo g N = lo g (M N)

requires both

M

and

N

to be positive. Going backwards,

lo g (M N)

can exist when both are negative —

lo g ((- 2) (- 8)) = lo g 16

is fine — while

lo g (- 2) + lo g (- 8)

does not exist at all. The law is safe left to right; going right to left, check the signs.

How do you prove a logarithmic identity or solve a logarithmic equation?

For a proof, take logarithms of the given relation and use the power law. For an equation, collapse both sides into single logarithms and then compare the insides — and always test the answer against the domain.

Worked proof 1. If

a^{2} = b^{3}

, prove that

lo g a = \frac{3}{2} lo g b

.

Take logs of both sides and apply the power law:

lo g a^{2} = lo g b^{3} ⟹ 2 lo g a = 3 lo g b ⟹ lo g a = \frac{3}{2} lo g b

The whole proof is the power law used once. Taking logs of a given equation is the standard opening move for this type.

Worked proof 2. If

a^{2} + b^{2} = 7 ab

, prove that

lo g \frac{a + b}{3} = \frac{1}{2} (lo g a + lo g b)

.

Start from the given relation and build a perfect square:

a^{2} + 2 ab + b^{2} = 7 ab + 2 ab = 9 ab ⟹ (a + b)^{2} = 9 ab ⟹ (\frac{a + b}{3})^{2} = ab

Now take logarithms of both sides:

2 lo g \frac{a + b}{3} = lo g (ab) = lo g a + lo g b

Dividing by

2

gives the required result.

Notice which chapter did the work. The algebra was the identity

(a + b)^{2} = a^{2} + 2 ab + b^{2}

from expansions; the logarithms only recorded the last step. Almost every proof of this shape is an expansions question with one log line at the end.

Worked equation 1. Solve

lo g (x + 1) + lo g (x - 1) = lo g 8

.

Collapse the left side, then compare:

lo g [(x + 1) (x - 1)] = lo g 8 ⟹ x^{2} - 1 = 8 ⟹ x^{2} = 9 ⟹ x = 3 or x = - 3

Now reject. If

x = - 3

then

x - 1 = - 4

, and the logarithm of a negative number does not exist. So

x = 3

only.

Check:

lo g 4 + lo g 2 = lo g 8

, as required.

Worked equation 2. Solve

lo g_{2} x + lo g_{2} (x - 2) = 3

lo g_{2} [x (x - 2)] = 3 ⟹ x (x - 2) = 2^{3} = 8 ⟹ x^{2} - 2 x - 8 = 0

Factorising by splitting the middle term,

(x - 4) (x + 2) = 0

, so

x = 4

x = - 2

. The value

- 2

makes both arguments negative, so

x = 4

.

Check:

lo g_{2} 4 + lo g_{2} 2 = 2 + 1 = 3

, as required.

This is the step that separates a full answer from a half one. A logarithmic equation almost always produces a root that has to be thrown away, and the rejection is worth a mark of its own. Solve, then check every argument is positive, then write the surviving answer.

Exam tip

What layout earns full marks on a logarithm question?

Name the law on every line, and put the domain check in writing at the end. These questions are short, so the examiner is reading your reasoning rather than your arithmetic.

- **Write by the product law or by the power law beside each step. It is one phrase and it protects you when a later line has a slip
- Expand in the order: quotient, product, power. Split the fraction before bringing indices down and no sign can go missing
- Keep brackets when a coefficient multiplies several terms.**

\frac{1}{2} (3 lo g x + lo g y - 5 lo g z)

without the bracket is a different expression
- Factorise numbers into primes before applying any law.

48

has to become

2^{4} \times 3

or the expansion cannot start
- When combining, push coefficients up as indices first, then read plus as multiply and minus as divide. Trying to do both at once is where terms get dropped
- For a proof, finish by writing the required statement, not the equivalent line you happen to have reached
- For an equation, always end with the rejection sentence: *

x = - 3

is rejected because

x - 1

would be negative.* A solved equation without it is an incomplete answer

The check that costs nothing. Substitute your answer into the original logarithms and confirm the arithmetic. In equation 2 above,

lo g_{2} 4 + lo g_{2} 2 = 3

takes five seconds and confirms both the root and the rejection at once.

Did you know

What happens if you take the logarithm of a logarithm's base?

The three laws all keep the base fixed. Here is a question they do not answer: what if you want to change it?

You can already handle a special case with nothing new. Consider

lo g_{2} 8

and

lo g_{8} 2

lo g_{2} 8 = 3 lo g_{8} 2 = \frac{1}{3}

They are reciprocals. Try another pair:

lo g_{9} 27 = \frac{3}{2}

from Part 1, and

lo g_{27} 9 = \frac{2}{3}

. Reciprocals again. In general

lo g_{a} b \times lo g_{b} a = 1

which says that swapping a logarithm's base and its argument turns the value upside down. You can prove it from Part 1's definition alone — write

lo g_{a} b = x

, so

a^{x} = b

, then raise both sides to the power

\frac{1}{x}

to get

a = b^{1/ x}

, which says

lo g_{b} a = \frac{1}{x}

.

That reciprocal is the first hint of the change-of-base rule, the Class 11 result that lets you rewrite any logarithm in terms of any base you like. It is the reason a calculator with only a base-ten button can still compute

lo g_{2} 1000

, and the reason the tables at the back of your textbook were printed in base ten and nothing else.

The underlying idea is worth keeping. In Part 1 the base looked like a fixed feature of the notation. It turns out to be one more thing that can be manipulated — and once it can, every logarithm in a problem can be brought onto common ground, exactly as you brought every index onto a common prime base.

Exam relevance

How are the logarithm laws used in JEE and NEET questions?

These three laws are foundation work that gets used, rather than tested, in competitive papers — which makes fluency more valuable than recall.

Where it leads in Mathematics. The Class 11 logarithm work assumes these laws completely and adds change of base; from there they run into Sequences and Series, Complex Numbers and then Calculus, where differentiating

lo g

of a product is done by expanding it with the product law first. JEE Main sets logarithmic equations and inequalities where the first two lines are exactly the collapsing you practised above.

Where it leads in Chemistry. The Nernst equation, pH calculations and first-order rate laws are all applications of these three laws, and both NEET and JEE Main ask numericals in which the chemistry is one step and the log manipulation is the rest. A candidate who can write

lo g \frac{[ A ] _{0}}{[ A ]}

as a difference in one move has an advantage over one who reaches for a calculator.

Question types to expect. Simplify a compound log expression to a number; solve an equation with logs on both sides; identify which option is valid. Assertion-reason items love the false laws — a statement like *

lo g (M + N) = lo g M + lo g N

* offered as the reason is a standard distractor.

The single trap that costs marks. Not rejecting the inadmissible root. Collapsing

lo g (x + 1) + lo g (x - 1)

into

lo g (x^{2} - 1)

widens the domain: the combined form is defined for

x < - 1

as well, but the original is not. Every root has to be tested in the original equation, and in JEE Advanced this is frequently the entire point of the question.

Board versus competitive emphasis. ICSE rewards the named law on each line and the closing rejection sentence. A competitive paper wants only the admissible root, but reaching it safely needs the same domain discipline. Build the habit while the marks are given for showing it.

Key takeaways

What should you be able to do with the logarithm laws before moving on?

Part 2 is three laws, each one a line of the indices chapter in different notation.

- **

lo g_{a} (M N) = lo g_{a} M + lo g_{a} N

lo g_{a} \frac{M}{N} = lo g_{a} M - lo g_{a} N

and

lo g_{a} M^{p} = p lo g_{a} M

, all needing positive arguments
- To expand, split the quotient, then the product, then bring indices down — and keep the bracket when a coefficient covers several terms
- To combine, push coefficients up as indices first, then read plus as multiply and minus as divide
- Steer the numbers towards powers of the base** and the answer becomes an integer:

lo g 2 + lo g 5 = 1

- Factorise into primes before applying any law to a numerical logarithm
- For a proof, take logs of the given relation, or build a perfect square first and take logs at the end
- For an equation, collapse, compare the insides, solve, and then reject any root that makes an argument zero or negative
- **There is still no law for

lo g (M + N)

**, for

\frac{l o g M}{l o g N}

, or for

(lo g M)^{2}

The fastest way to find out whether this has stuck is to expand

lo g \frac{x ^{3} y}{z ^{5}}

and then solve

lo g_{2} x + lo g_{2} (x - 2) = 3

without looking back — and to notice whether you remembered the rejection line on your own.

Why do logarithms turn multiplication into addition?

What are the three laws of logarithms and where does each come from?

How do you expand a logarithm of a messy fraction into separate terms?

How do you collapse a sum of logarithms into one and evaluate it?

How do you prove a logarithmic identity or solve a logarithmic equation?

What layout earns full marks on a logarithm question?

What happens if you take the logarithm of a logarithm's base?

How are the logarithm laws used in JEE and NEET questions?

What should you be able to do with the logarithm laws before moving on?

Related study notes

Ready to put this into practice?