Exponentials & Logarithms | The Maths Tailor — UK Methods

Exponentials & Logarithms

7 methods · 19 questions

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 $u$ 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 (7)

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}

Voir la correction

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

or

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{l o g _{10} 6}{2 l o g _{10} 2}

. Therefore

p - q = \frac{l o g _{10} 6}{2 l o g _{10} 2} - \frac{1}{2} = \frac{l o g _{10} 6 - l o g _{10} 2}{2 l o g _{10} 2} = \frac{l o g _{10} 3}{2 l o g _{10} 2} = \frac{l o g _{10} 3}{l o 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

Voir la correction

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

or

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

Voir la correction

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

or

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.

TMUA 2020 Paper 1 — Q6

Find the maximum value of the function

f (x) = \frac{1}{5 ^{2 x} - 4 ( 5 ^{x} ) + 7}

(A)

\frac{1}{7}

(B)

\frac{1}{4}

(C)

\frac{1}{3}

(D)

3

(E)

4

(F)

7

Voir la correction

Let

u = 5^{x} > 0

. The denominator becomes

u^{2} - 4 u + 7 = (u - 2)^{2} + 3

. This is minimised when

u = 2

(which is achievable since

5^{x} = 2

has a real solution), giving minimum denominator value

3

. The fraction

\frac{1}{denominator}

is therefore maximised at

\frac{1}{3}

. Hence the answer is C.

TMUA 2021 Paper 1 — Q4

Find the minimum value of the function

\frac{1}{denominator}

(A)

- 16

(B)

- 12

(C)

- 8

(D)

0

(E)

4

(F)

20

Voir la correction

Let

u = 2^{x}

. The function becomes

u^{2} - 2^{3} u + 4 = u^{2} - 8 u + 4

. Completing the square:

(u - 4)^{2} - 16 + 4 = (u - 4)^{2} - 12

. The minimum is

- 12

, attained when

u = 4

, i.e.

2^{x} = 4

, i.e.

x = 2

, which is a valid real value. Hence the answer is B.

TMUA 2023 Paper 1 — Q7

P (x)

and

Q (x)

are defined as follows:

2^{x} = 4

Q (x) = 2^{(2 x - 2)} - 2^{(x + 2)} + 16

Find the largest value of

x

such that

P (x)

and

Q (x)

are in the ratio

4 : 1

, respectively.

(A)

5

(B)

12

(C)

32

(D)

lo g_{2} 3

(E)

lo g_{2} 5

(F)

lo g_{2} 12

(G)

lo g_{2} 20

Voir la correction

The condition

P (x) : Q (x) = 4 : 1

means

P (x) = 4 Q (x)

:

2^{x} + 4 = 4 (2^{2 x - 2} - 2^{x + 2} + 16) = 2^{2 x} - 2^{x + 4} + 64.

Let

u = 2^{x}

. Then:

u + 4 = u^{2} - 16 u + 64

u = 2^{x}

(u - 5) (u - 12) = 0

So

u = 5

or

u = 12

, giving

x = lo g_{2} 5

or

x = lo g_{2} 12

.

The largest value is

lo g_{2} 12

. Hence the answer is F.

TMUA Specimen Paper 1 — Q11

The sum of the roots of the equation

x = lo g_{2} 12

is

(A)

3

(B)

8

(C)

2 lo g_{10} 2

(D)

lo g_{10} (\frac{15}{4})

(E)

\frac{lo g _{10} 15}{lo g _{10} 2}

Voir la correction

Let

y = 2^{x}

. The equation becomes

y^{2} - 8 y + 15 = 0

, which factorises as

(y - 3) (y - 5) = 0

, giving

y = 3

or

y = 5

, so

2^{x} = 3

or

2^{x} = 5

. Taking

lo g_{10}

: the first root is

x_{1} = \frac{lo g _{10} 3}{lo g _{10} 2}

and the second is

x_{2} = \frac{lo g _{10} 5}{lo g _{10} 2}

. Their sum is

x_{1} + x_{2} = \frac{lo g _{10} 3 + lo g _{10} 5}{lo g _{10} 2} = \frac{lo g _{10} 15}{lo g _{10} 2}

. Hence the answer is E.