CDS Mathematics · Surds, Indices and Simplification

Exponential Equations

An equation in a^x becomes a quadratic when you put t = a^x, and a sum of like powers becomes one power once the common factor is taken out.

All Surds, Indices and Simplification notes CDS Mathematics strategy

Why this matters

Eight PYQs, all MODERATE. Two moves solve every one: substitute t for the power (a^x + a^(−x) becomes t + 1/t), or take out the smallest power as a common factor.

Concept 1 of 2: Substitute t = a^x

a^{1 + x}

and

a^{1 - x}

are

a\cdot t

and

\dfrac at

when

t = a^x

. Multiply through by

t

and the equation is a quadratic in

t

; each positive root gives one value of

x

Definition

Put $t = a^x$ (so $t > 0$ ): $a^{x + k} = a^k t$ and $a^{-x} = \dfrac1t$ .
Solve the quadratic in $t$ , reject any $t \le 0$ , then $x = \log_a t$ .
$t + \dfrac1t = 2$ forces $t = 1$ , so $x = 0$ ; $t + \dfrac1t > 2$ for every other positive $t$ .
If $pq = 1$ (like $(2 + \sqrt3)(2 - \sqrt3)$ ), then $q^x = \dfrac{1}{p^x}$ and the same substitution works.

Quadratic in t

a^{1 + x} + a^{1 - x} = c \;\Rightarrow\; a t^2 - c t + a = 0,\ \ t = a^x

Worked example

Solve

2^{x + 1} + 2^{3 - x} = 17

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2017 · CDS (I) 2017 — Elementary Mathematics · Q29Moderate

Example 1 · Surds, Indices and Simplification · Exponential Equations

The values of

x

which satisfy the equation

5^{1+x} + 5^{1-x} = 26

are

Reject non-positive t

A power

a^x

with

a > 0

is always positive. A root

t = -2

of the quadratic gives no

x

, and counting it doubles the answer.

Drill 5 more on substitute t = a^x

Concept 2 of 2: Take out the common power

5^{x + 1} - 5^{x - 1}

5^{x - 1}

times

(5^2 - 1)

. Taking out the smallest power turns a sum of powers into one power times a number.

Definition

$a^{x + m} \pm a^{x + n} = a^{x + n}(a^{m - n} \pm 1)$ for $m > n$ .
Write every number as a power of one prime where possible: $2187 = 3^7$ , $4^{xy} = 2^{2xy}$ .
Two equations in the exponents: compare exponents of the same base, then solve the pair.

Common factor

a^{x + 1} - a^{x - 1} = a^{x - 1}(a^2 - 1)

Worked example

3^{x + 2} - 3^{x} = 216

, find

x

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2020 · CDS (I) 2020 — Elementary Mathematics · Q15Moderate

Example 2 · Surds, Indices and Simplification · Exponential Equations

5^{x+1} - 5^{x-1} = 600

, then what is the value of

10^{2x}

Answer what is asked

Solving for

x

is usually only the first step: the question may want

10^{2x}

x + y

. Keep the requested expression in view.

Drill 1 more on take out the common power

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 (2)

Substitute t = a^x
Quadratic in t
$a^{1 + x} + a^{1 - x} = c \;\Rightarrow\; a t^2 - c t + a = 0,\ \ t = a^x$
Take out the common power
Common factor
$a^{x + 1} - a^{x - 1} = a^{x - 1}(a^2 - 1)$

Watch out for (2)

Reject non-positive t
→ Substitute t = a^x
Answer what is asked
→ Take out the common power

Test yourself on Surds, Indices and Simplification

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