Exponentials & Logarithms

9 methods · 33 questions

★ Signature method unlocked: the technique this chapter rewards most often, with 3 worked questions.

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Whenever an equation is built from a single repeated exponential block (the same $a^{x}$ appearing as $a^{x}$ , $a^{2 x}$ , $a^{3 x}$ , a constant, ...), set $u$ equal to that block. Every term becomes a power of $u$ , the equation collapses to an ordinary polynomial (usually a quadratic), you solve for $u$ , then read off $x$ from $u = (base)^{x}$ .

Trigger: the variable appears only inside powers of one base, with exponents that are integer multiples of a single $x$ -term.

Instances:
(i) a plain $a^{2 x}$ and $a^{x}$ pattern gives a quadratic in $u = a^{x}$ ;
(ii) after rewriting to one base, $a^{2 x}$ becomes $u^{2}$ so even $4^{2 x}$ or $9^{x}$ fits;
(iii) a fractional exponent like $3^{x /2}$ is itself the right $u$ , with $3^{x} = u^{2}$ .

Linked questions (3)

TMUA 2016 Paper 1 · Q11

The real roots of the equation

4^{2 x} + 12 = 2^{2 x + 3}

are

p

and

q

, where

p > q

. The value of

p - q

can be expressed as

A $\frac{3}{4}$
B $1$
C $4$
D $- \frac{1}{2} + lo g_{10} \frac{3}{2}$
E $\frac{lo g _{10} 3}{lo g _{10} 4}$
F $\frac{lo g _{10} 3}{lo g _{10} 2}$

Worked solution

Answer:E

Note that

4^{2 x} = (2^{2})^{2 x} = 2^{4 x} = (2^{2 x})^{2}

and

2^{2 x + 3} = 8 \cdot 2^{2 x}

. Let

y = 2^{2 x}

; the equation becomes

y^{2} + 12 = 8 y

, i.e.

y^{2} - 8 y + 12 = 0

, which factorises as

(y - 6) (y - 2) = 0

, giving

y = 6

y = 2

. Since

p > q

, we have

2^{2 p} = 6

and

2^{2 q} = 2

. From the second:

2 q = 1

, so

q = \frac{1}{2}

. From the first:

2 p lo g_{10} 2 = lo g_{10} 6

, so

p = \frac{lo g _{10} 6}{2 lo g _{10} 2}

. Therefore

p - q = \frac{lo g _{10} 6}{2 lo g _{10} 2} - \frac{1}{2} = \frac{lo g _{10} 6 - lo g _{10} 2}{2 lo g _{10} 2} = \frac{lo g _{10} 3}{2 lo g _{10} 2} = \frac{lo g _{10} 3}{lo g _{10} 4}

. Hence the answer is E.

TMUA 2018 Paper 1 · Q15

Find the sum of the real solutions of the equation:

3^{x} - (3)^{x + 4} + 20 = 0

A $1$
B $4$
C $9$
D $lo g_{3} 20$
E $2 lo g_{3} 20$
F $4 lo g_{3} 20$

Worked solution

Answer:E

Note that

(3)^{x + 4} = 3^{(x + 4) /2} = 3^{x /2} \cdot 3^{2} = 9 \cdot 3^{x /2}

. Let

y = 3^{x /2}

(so

y > 0

and

3^{x} = y^{2}

). The equation becomes:

y^{2} - 9 y + 20 = 0 ⟹ (y - 4) (y - 5) = 0,

giving

y = 4

y = 5

. For

y = 4

3^{x /2} = 4 ⟹ \frac{x}{2} = lo g_{3} 4 ⟹ x = 2 lo g_{3} 4

. For

y = 5

3^{x /2} = 5 ⟹ x = 2 lo g_{3} 5

. Both solutions are real. Their sum is

2 lo g_{3} 4 + 2 lo g_{3} 5 = 2 lo g_{3} 20

. Hence the answer is E.

TMUA 2019 Paper 1 · Q15

Find the real non-zero solution to the equation

\frac{2 ^{(9^{x})}}{8 ^{(3^{x})}} = \frac{1}{4}

A $lo g_{3} 2$
B $2 lo g_{3} 2$
C $1$
D $2$
E $lo g_{2} 3$
F $2 lo g_{2} 3$

Worked solution

Answer:A

Let

u = 3^{x}

, so

9^{x} = (3^{2})^{x} = (3^{x})^{2} = u^{2}

and

8^{3^{x}} = 2^{3 u}

. The equation becomes

\frac{2 ^{u^{2}}}{2 ^{3 u}} = 2^{- 2}

, so

u^{2} - 3 u = - 2

, giving

u^{2} - 3 u + 2 = 0

, i.e.

(u - 1) (u - 2) = 0

. Thus

u = 1

u = 2

, meaning

3^{x} = 1

(giving

x = 0

) or

3^{x} = 2

(giving

x = lo g_{3} 2

). The non-zero solution is

x = lo g_{3} 2

. Hence the answer is A.