NDA Maths · Logarithms

Solving Logarithmic Equations & Applications

Solving a log equation means turning it into an algebraic one — take the log of an exponential equation, substitute t = aˣ to reach a quadratic, then reject any root that breaks a domain.

All Logarithms notes NDA Maths strategy

Why this matters

Eleven PYQs, three of them HARD — this is where the chapter's difficulty concentrates. The recurring moves are few: collapse both sides to a single log and drop the log, substitute t = aˣ to get a quadratic, or take log₁₀ of an exponential equation. The HARD ones add a twist — a GP/chain-rule condition or an AM-GM bound — but the spine is always the same. Domain-checking is what separates a 4-mark answer from a wrong one.

Test yourself — a quick 15-question Logarithms recall quiz, instant score.

Concept 1 of 5: Taking the Log of an Exponential Equation

When the unknown sits in an exponent —

a^x = b

— apply a log to both sides to bring the exponent down:

x\log a = \log b

. With a supplied value of

\log_{10} 2

you can then compute a decimal, or you can just collapse log expressions of the form

k\,f(\cdot)

until they reduce to

\log 10 = 1

Definition

Core move: $a^x = b \Rightarrow x = \dfrac{\log b}{\log a}$ (any common base). Two flavours appear:

Numeric: given $\log_{10} 2$ , evaluate $x$ from $(0.2)^x = 2$ by writing $\log_{10} 0.2 = \log_{10}\tfrac{2}{10} = \log_{10} 2 - 1$ .
Collapse-to-1: expressions built from $\log_{10} 2$ and $\log_{10} 5$ simplify because $\log_{10} 2 + \log_{10} 5 = \log_{10} 10 = 1$ . Group terms to expose a $\log_{10}(\text{product} = 10^k)$ .

Bring the exponent down with a log

a^x = b \;\Rightarrow\; x = \dfrac{\log b}{\log a}

Worked example

3^x = 5

and

\log_{10} 3 = 0.477,\ \log_{10} 5 = 0.699

, find

x

to two decimals.

Practice this conceptself-check

The same idea in a real exam question:

NDA · Sep 2018 · Q31Moderate

Example 1 · Logarithms · Solving Logarithmic Equations and Applications

(0.2)^x = 2

and

\log_{10} 2 = 0.3010

, then what is the value of x to the nearest tenth?

$\log_{10} 0.2 = \log_{10} 2 - 1$ , which is negative

Writing

0.2 = \tfrac{2}{10}

gives

\log_{10} 0.2 = \log_{10} 2 - \log_{10} 10 = 0.3010 - 1 = -0.6990

. Dropping the

-1

flips the sign of

x

— the answer to

(0.2)^x = 2

is negative.

Drill 2 more on taking the log of an exponential equation

Concept 2 of 5: Substitution t = aˣ to a Quadratic

When a log equation mixes

2^x

and

2^{2x}

(or

3^x

and

9^x

), collapse both sides to a single log, drop the log, then let

t = 2^x

. The relation becomes a quadratic in

t

; solve it and discard any non-positive

t

, because

2^x

is always positive.

Definition

Recipe:

Collapse to one log on each side using the laws, then equate arguments (since $\log_a M = \log_a N \Rightarrow M = N$ ).
Substitute $t = a^x$ (so $a^{2x} = t^2$ ). The equation becomes a quadratic $t^2 + bt + c = 0$ .
Solve and screen: factorise ( $t^2-5t+4=(t-1)(t-4)$ ) or use $t=\dfrac{-b\pm\sqrt{b^2-4c}}{2}$ , then reject any root $t \le 0$ — an exponential $a^x$ can never be $0$ or negative. From the surviving $t$ , recover $x = \log_a t$ .

An AP condition on three logs, $2\log(2^x-1) = \log 2 + \log(2^x+3)$ , feeds straight into this: square out to $(2^x-1)^2 = 2(2^x+3)$ , a quadratic in $t = 2^x$ .

Let t = aˣ and solve the quadratic (keep t > 0)

t = a^x > 0,\quad t^2 + bt + c = 0 \;\Rightarrow\; x = \log_a t

Worked example

Solve

4^x - 3\cdot 2^x - 4 = 0

Practice this conceptself-check

The same idea in a real exam question:

NDA · Sep 2017 · Q1Moderate

Example 2 · Logarithms · Solving Logarithmic Equations and Applications

x + \log_{10}(1 + 2^x) = x\log_{10} 5 + \log_{10} 6

, then

x

is equal to

Throw out the non-positive t

A quadratic in

t = 2^x

often hands you a negative root. Because

2^x

is strictly positive, that root yields no real

x

— keep only

t > 0

before solving

x = \log_2 t

Drill 2 more on substitution t = aˣ to a quadratic

Concept 3 of 5: Domain Checks & Counting Solutions

A log equation with two different bases (

\log_4

and

\log_2

) is solved by bringing both to one base, but the algebra can hand you roots that make some argument non-positive. The genuine count of solutions is only the roots that survive the domain — argument

>0

for every log in the original equation.

Definition

Procedure for "how many solutions?":

Unify the base: $\log_4(x-1) = \tfrac{1}{2}\log_2(x-1)$ , so $\log_4(x-1) = \log_2(x-3)$ becomes $\tfrac12\log_2(x-1) = \log_2(x-3)\Rightarrow x-1 = (x-3)^2$ .
Solve the resulting polynomial, then impose the domain: every original argument must be $>0$ (here $x-1>0$ AND $x-3>0$ , so $x>3$ ).
Inequalities: for $x^{\log_7 x} > 7$ , take $\log_7$ of both sides to get $(\log_7 x)^2 > 1$ , then $|\log_7 x| > 1$ gives $x > 7$ or $0 < x < \tfrac17$ .

Keep only roots with every argument > 0

\log_a M = \log_a N \Rightarrow M = N,\ \text{then require } M,N > 0

Worked example

How many solutions does

\log_9(x+6) = \log_3 x

have?

The same idea in a real exam question:

NDA · Apr 2024 · Q27Moderate

Example 3 · Logarithms · Solving Logarithmic Equations and Applications

What is the number of solutions of

\log_4(x-1)=\log_2(x-3)

An algebraic root is not a solution until the domain clears it

x-1 = (x-3)^2

gives

x = 2

and

x = 5

, but

x = 2

makes

x-3 = -1 < 0

, so

\log_2(x-3)

is undefined. Only

x = 5

is a genuine solution — always re-substitute into the ORIGINAL equation.

The argument must be strictly positive — $0$ is not allowed

\log_a x

is undefined for

x \le 0

, and that includes

x = 0

: there is no power of

a

that gives

0

. A root that makes ANY argument equal to

0

(not just negative) must be rejected — the domain requires every argument

> 0

, strictly.

Drill 1 more on domain checks & counting solutions

Concept 4 of 5: GP, Chain-Rule & AM-GM Conditions

The HARD log questions wrap a sequence or inequality condition around the same tools. A GP condition multiplies the outer terms; the chain rule

\log_x a\cdot\log_b x = \log_b a

collapses the product; and a "

k

can never equal" question is usually AM-GM,

t + \tfrac1t \ge 2

, in disguise.

Definition

Two HARD patterns:

GP of logs: $\log_x a,\ a^x,\ \log_b x$ in GP means $(a^x)^2 = \log_x a\cdot\log_b x$ . The product collapses by the chain rule $\log_x a\cdot\log_b x = \log_b a$ , giving $a^{2x} = \log_b a$ ; take $\log_a$ to solve for $x$ .
AM-GM bound: writing $\log_x\tfrac{x}{y} + \log_y\tfrac{y}{x}$ with $t = \log_x y \ge 1$ gives $2 - t - \tfrac1t$ . Since $t + \tfrac1t \ge 2$ (AM-GM), the expression is $\le 0$ — so it can never equal any positive value.

Chain rule and the AM-GM floor

\log_x a\cdot\log_b x = \log_b a, \qquad t + \tfrac{1}{t} \ge 2\ (t>0)

Worked example

Find the maximum value of

2 - \Big(t + \dfrac{1}{t}\Big)

for

t > 0

The same idea in a real exam question:

NDA · Apr 2024 · Q28Hard

Example 4 · Logarithms · Solving Logarithmic Equations and Applications

For

x\geq y>1

, let

\log_x\!\dfrac{x}{y}+\log_y\!\dfrac{y}{x}=k

, then the value of

k

can never be equal to

The GP condition squares the MIDDLE term

For

p, q, r

in GP the relation is

q^2 = pr

— the middle term is squared and set equal to the product of the outer two. Squaring the wrong term derails the whole chain-rule collapse.

Drill 1 more on gp, chain-rule & am-gm conditions

Concept 5 of 5: Application — Trailing Zeros of a Factorial

The number of trailing zeros of

n!

is set by how many times

10 = 2\times 5

divides it — and since factors of

5

are scarcer than factors of

2

, you just count the

5

s. This is the chapter's one number-theory application that rides on the same "count the powers" instinct logs train.

Definition

Legendre's count (for the prime $5$ ):

Z(n) = \left\lfloor\dfrac{n}{5}\right\rfloor + \left\lfloor\dfrac{n}{25}\right\rfloor + \left\lfloor\dfrac{n}{125}\right\rfloor + \cdots

gives the number of trailing zeros of

n!

(the power of

5

n!

; the power of

2

is always larger, so

5

is the bottleneck). To find how many $n$ give exactly $z$ zeros, note

Z(n)

is non-decreasing and jumps at multiples of

5

: a run of consecutive integers shares the same

Z

value until the next multiple of

5

Legendre — trailing zeros count the 5s

Z(n) = \sum_{i\ge 1}\left\lfloor \dfrac{n}{5^i} \right\rfloor

Worked example

How many trailing zeros does

30!

have?

Practice this concept2 quick reps

The same idea in a real exam question:

NDA · Sep 2025 · Q4Moderate

Example 5 · Logarithms · Solving Logarithmic Equations and Applications

Let

n

be a natural number. The number of consecutive zeros at the end of the expansion of

n!

is exactly 2. How many values of

n

are possible?

Count the 5s, not the 2s — and don't forget 25, 125…

Each multiple of

25

contributes an EXTRA

5

beyond the one already counted by

\lfloor n/5\rfloor

. Stopping at

\lfloor n/5\rfloor

undercounts for

n \ge 25

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (5)

Taking the Log of an Exponential Equation
Bring the exponent down with a log
$a^x = b \;\Rightarrow\; x = \dfrac{\log b}{\log a}$
Substitution t = aˣ to a Quadratic
Let t = aˣ and solve the quadratic (keep t > 0)
$t = a^x > 0,\quad t^2 + bt + c = 0 \;\Rightarrow\; x = \log_a t$
Domain Checks & Counting Solutions
Keep only roots with every argument > 0
$\log_a M = \log_a N \Rightarrow M = N,\ \text{then require } M,N > 0$
GP, Chain-Rule & AM-GM Conditions
Chain rule and the AM-GM floor
$\log_x a\cdot\log_b x = \log_b a, \qquad t + \tfrac{1}{t} \ge 2\ (t>0)$
Application — Trailing Zeros of a Factorial
Legendre — trailing zeros count the 5s
$Z(n) = \sum_{i\ge 1}\left\lfloor \dfrac{n}{5^i} \right\rfloor$

Watch out for (6)

$\log_{10} 0.2 = \log_{10} 2 - 1$ , which is negative
→ Taking the Log of an Exponential Equation
Throw out the non-positive t
→ Substitution t = aˣ to a Quadratic
An algebraic root is not a solution until the domain clears it
→ Domain Checks & Counting Solutions
The argument must be strictly positive — $0$ is not allowed
→ Domain Checks & Counting Solutions
The GP condition squares the MIDDLE term
→ GP, Chain-Rule & AM-GM Conditions
Count the 5s, not the 2s — and don't forget 25, 125…
→ Application — Trailing Zeros of a Factorial

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