Exponentials & Logarithms

6 methods · 9 questions

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Whenever an equation is built from different powers of a single exponential block or a single logarithm, set $u$ equal to that block. The equation collapses to an ordinary polynomial (usually a quadratic). Solve for $u$ , then read off $x$ .

Trigger: The variable appears inside powers of one exponential base, or as powers of a logarithm like $(lo g x)^{2}$ .

Instances:
(i) for exponentials, split additions in the exponent first ( $a^{2 x + 1} = a (a^{x})^{2}$ ) and reject any $u \leq 0$ since $a^{x} > 0$ ;
(ii) for logarithms, pull exponents out of arguments first ( $lo g (x^{2}) = 2 lo g x$ ) before substituting $u = lo g x$ ;
(iii) for bases like $2$ , setting $u = (2)^{x}$ makes $2^{x} = u^{2}$ .

Linked questions (3)

NSAA 2019 Section 1, Part E (Mathematics questions) · Q81

Which one of the following is the real solution of the equation

3 \times 5^{2 x + 1} - 5^{x} - 2 = 0

A $x = lo g_{5} (\frac{1}{3})$
B $x = lo g_{5} (\frac{2}{5})$
C $x = lo g_{5} (\frac{3}{5})$
D $x = lo g_{5} (\frac{2}{3})$
E $x = lo g_{5} (\frac{5}{3})$
F $x = lo g_{5} (\frac{5}{2})$

Worked solution

Answer:B

Substitute

u = 5^{x}

, remembering that

u > 0

for every real

x

. Since

5^{2 x + 1} = 5 \times 5^{2 x} = 5 (5^{x})^{2} = 5 u^{2},

the equation becomes

15 u^{2} - u - 2 = 0.

Two numbers with product

15 \times (- 2) = - 30

and sum

- 1

are

- 6

and

5

, so

15 u^{2} - 6 u + 5 u - 2 = 3 u (5 u - 2) + (5 u - 2) = (5 u - 2) (3 u + 1) = 0.

Hence

u = \frac{2}{5}

u = - \frac{1}{3}

. A power of

5

is always positive, so the negative root is rejected and there is exactly one real solution.

From

5^{x} = \frac{2}{5}

we get

x = lo g_{5} (\frac{2}{5}) .

The answer is B.

NSAA 2017 Section 1, Part E (Mathematics questions) · Q77

Which of the following is a solution to the equation

3^{(2 x + 1)} - 6 (3^{x}) = 0

A $lo g_{2} 3$
B $lo g_{3} 2$
C $2$
D $lo g_{10} 2$
E $\frac{2}{3}$

Worked solution

Answer:B

Split the first power so that both terms are built from

3^{x}

3^{2 x + 1} = 3 \times 3^{2 x} = 3 (3^{x})^{2} .

Substituting

t = 3^{x}

, which is strictly positive for every real

x

, turns the equation into

3 t^{2} - 6 t = 0, that is 3 t (t - 2) = 0.

The root

t = 0

is impossible because

3^{x}

is never zero, so

t = 2

, that is

3^{x} = 2

.

Taking logarithms to base

3

gives

x = lo g_{3} 2

.

As a check,

3^{x} = 2

makes the first term

3 \times 2^{2} = 12

and the second term

6 \times 2 = 12

, so their difference is zero.

The answer is B.

ENGAA 2020 Section 1, Part B (Mathematics questions) · Q37

Find the product of the real roots of the equation

(lo g_{10} x^{2})^{2} + lo g_{10} x = 3

A $1 0^{- \frac{3}{2}}$
B $1 0^{- 1}$
C $1 0^{- \frac{1}{2}}$
D $1 0^{- \frac{1}{4}}$
E $1 0^{\frac{3}{5}}$
F $1 0^{1}$

Worked solution

Answer:D

The term

lo g_{10} x

requires

x > 0

, and on that domain

lo g_{10} x^{2} = 2 lo g_{10} x

.

Put

u = lo g_{10} x

. The equation becomes

(2 u)^{2} + u = 3 ⟹ 4 u^{2} + u - 3 = 0.

This factorises as

(4 u - 3) (u + 1) = 0

, so

u = \frac{3}{4}

u = - 1

, which gives the two real roots

x = 1 0^{\frac{3}{4}}

and

x = 1 0^{- 1}

.

Their product is

1 0^{\frac{3}{4}} \times 1 0^{- 1} = 1 0^{\frac{3}{4} - 1} = 1 0^{- \frac{1}{4}} .

The answer is D.