Number & Units

7 methods · 15 questions

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When an equation involves powers of different but related bases, rewrite every base as a power of their common prime base before combining them. Do not attempt to multiply or divide terms with different bases directly.

Trigger: An equation or expression with indices where the bases are all powers of 2 (like 2, 4, 8) or powers of 3 (like 3, 9, 27, 81).

Instances:
(i) Rewrite each base: $4^{x} = (2^{2})^{x} = 2^{2 x}$ and $2 7^{x} = (3^{3})^{x} = 3^{3 x}$ ;
(ii) Use index laws to combine into a single power on each side;
(iii) Equate the exponents: if $a^{x} = a^{y}$ and $a > 0$ , $a \neq = 1$ , then $x = y$ .

Linked questions (3)

NSAA 2019 Section 1, Part A (Mathematics questions) · Q9

Given that

9^{2 x - 1} \times \frac{1}{2 7 ^{x}} = 8 1^{x}

what is the value of

x

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

Worked solution

Answer:A

Every base here is a power of

3

9 = 3^{2}

27 = 3^{3}

and

81 = 3^{4}

. Rewriting the left-hand side,

(3^{2})^{2 x - 1} \times (3^{3})^{- x} = 3^{2 (2 x - 1)} \times 3^{- 3 x} = 3^{4 x - 2 - 3 x} = 3^{x - 2},

while the right-hand side is

(3^{4})^{x} = 3^{4 x}

.

Since

t \mapsto 3^{t}

is one-to-one,

3^{x - 2} = 3^{4 x}

forces

x - 2 = 4 x

, so

- 2 = 3 x

and

x = - \frac{2}{3} .

The answer is A.

NSAA 2017 Section 1, Part A (Mathematics questions) · Q9

What is the value of

x

that makes the following expression correct?

2^{3 + 2 x} 4^{x} 8^{- x} = 42

A $- 2.25$
B $- 1.75$
C $- 1.5$
D $- 0.5$
E $- 0.25$

Worked solution

Answer:D

Write every factor as a power of the same base. Since

4 = 2^{2}

and

8 = 2^{3}

4^{x} = 2^{2 x}, 8^{- x} = 2^{- 3 x} .

The left-hand side then collapses to a single power:

2^{3 + 2 x} \times 2^{2 x} \times 2^{- 3 x} = 2^{3 + 2 x + 2 x - 3 x} = 2^{3 + x} .

The right-hand side is

42 = 2^{2} \times 2^{1/2} = 2^{5/2}

.

Powers of

2

are equal only when their exponents are equal, so

3 + x = \frac{5}{2}

and

x = \frac{5}{2} - 3 = - \frac{1}{2} = - 0.5.

The answer is D.

NSAA 2016 Section 1, Part A (Mathematics questions) · Q13

Which one of the following expressions is equivalent to

\frac{9 ^{2 n + 1} \times 3 ^{4 - 3 n}}{2 7 ^{2 - n}}

A $3^{9}$
B $3^{- 2 n}$
C $3^{2 - 2 n}$
D $3^{4 n}$
E $3^{6 n - 2}$
F $3^{6}$

Worked solution

Answer:D

Write every base as a power of

3

, using

9 = 3^{2}

and

27 = 3^{3}

9^{2 n + 1} = 3^{2 (2 n + 1)} = 3^{4 n + 2}, 2 7^{2 - n} = 3^{3 (2 - n)} = 3^{6 - 3 n} .

Multiplying powers of the same base adds the exponents, so the numerator is

3^{4 n + 2} \times 3^{4 - 3 n} = 3^{(4 n + 2) + (4 - 3 n)} = 3^{n + 6} .

Dividing subtracts the exponents:

\frac{3 ^{n + 6}}{3 ^{6 - 3 n}} = 3^{(n + 6) - (6 - 3 n)} = 3^{4 n} .

As a check,

n = 1

gives

\frac{9 ^{3} \times 3}{27} = \frac{729 \times 3}{27} = 81 = 3^{4}

.

The answer is D.