Exam practice

$\dfrac{1}{343}, \dfrac{1}{49}, \dfrac17, 1, 7, 49, 343$

$125, 25, 5, 1, \dfrac15, \dfrac{1}{25}, \dfrac{1}{125}$. ($5^{-x}$ gets smaller as x grows: $5^{-x} = \left(\dfrac15\right)^x$.)

$3^x = 3^2$: x = 2.

$5^{x - 1} = 5^2$, x − 1 = 2: x = 3.

$5^{-(2x - 4)} = 5^3$, so −2x + 4 = 3: $x = \dfrac12$. Check: $5^{2x - 4} = 5^{-3}$, and $\dfrac{1}{5^{-3}} = 125$ ✓.

$0.125 = \dfrac18 = 2^{-3}$ and $4^{x - 2} = 2^{2x - 4}$, so 2x − 4 = −3: $x = \dfrac12$.

$1\dfrac23 = \dfrac53 = \left(\dfrac35\right)^{-1}$, so $\left(\dfrac35\right)^x = \left(\dfrac35\right)^{-3}$: x = −3.

$3^{x + 1} = 3 \times 3^x$, so $3 \times 2^x \times 3^x = 18$, $6^x = 6$: x = 1.

$\dfrac{1 - x}{1 + x} = \dfrac13$, so 3 − 3x = 1 + x, 4x = 2: $x = \dfrac12$.

Write both sides as powers of 2: a root is a fractional index.

$$2^{\frac{2(x + 8)}{2x + 4}} = 2^{\frac76}$$

$\dfrac{x + 8}{x + 2} = \dfrac76$, so 6x + 48 = 7x + 14: x = 34.

$2^x(2 + 4 + 8) = 448$, $2^x = 32 = 2^5$: x = 5.

$3^x(3 - 1) = 162$, $3^x = 81 = 3^4$: x = 4.

$4^{x - 1} = \dfrac{4^x}{4}$, so $4^x(4 - 2) = 32$, $4^x = 16 = 4^2$: x = 2.

$$3^x\left(12 - 9 - \frac13\right) = 72$$

$3^x \times \dfrac83 = 72$, $3^x = 27 = 3^3$: x = 3.

$3^x(9 + 3 + 2) = 126$, $3^x = 9$: x = 2.

Put the powers of 2 on one side and the powers of 3 on the other.

$$\begin{aligned} 2^x + 2 \times 2^x &= 3^x - \frac{3^x}{9} \\ 3 \times 2^x &= \frac89 \times 3^x \\ \frac{2^x}{3^x} &= \frac{8}{27} \\ \left(\frac23\right)^x &= \left(\frac23\right)^3 \end{aligned}$$

x = 3.

$8^{x - 1} = \dfrac{2^{3x}}{8}$ and $2^{3x - 2} = \dfrac{2^{3x}}{4}$, so $2^{3x}\left(\dfrac18 - \dfrac14\right) = -8$, $2^{3x} \times \left(-\dfrac18\right) = -8$, $2^{3x} = 64 = 2^6$: x = 2.

Let $s = \sqrt{5x + 1}$ and write both sides as powers of 2.

With $s = \sqrt{5x + 1}$: $2^{-2(2 - s)} = 2^{2 + s}$, so 2s − 4 = 2 + s, s = 6.
5x + 1 = 36: x = 7.

$5^x + \dfrac{1}{5^x} = 5 + \dfrac15$, so $5^x = 5$ or $\dfrac15$ (let $5^x = a$: $5a^2 - 26a + 5 = 0$, (a − 5)(5a − 1) = 0). x = ±1.

Let $7^x = a$: $7a^2 - 50a + 7 = 0$, (a − 7)(7a − 1) = 0, a = 7 or $\dfrac17$. x = ±1.

Let $9^x = a$: $9a^2 - 82a + 9 = 0$, (a − 9)(9a − 1) = 0. x = ±1.

Let $4^x = a$: $16a^2 - 257a + 16 = 0$, (a − 16)(16a − 1) = 0, a = 16 or $\dfrac{1}{16}$. x = ±2.

Let $5^x = a$: $25a^2 - 626a + 25 = 0$, (a − 25)(25a − 1) = 0. x = ±2.

Let $3^x = a$: $81a + \dfrac1a = 30$, so $81a^2 - 30a + 1 = 0$, (27a − 1)(3a − 1) = 0, $a = \dfrac{1}{27}$ or $\dfrac13$.
x = −3 or −1. Check x = −1: $81 \times \dfrac13 + 3 = 30$ ✓.

Let $3^x = a$: $12a - a^2 = 27$, $a^2 - 12a + 27 = 0$, (a − 3)(a − 9) = 0. x = 1 or 2.

Let $2^p = a$: $6a - a^2 = 8$, (a − 2)(a − 4) = 0. p = 1 or 2.

Let $5^x = a$: $a^2 - 30a + 125 = 0$, (a − 5)(a − 25) = 0. x = 1 or 2.

Let $2^x = a$: $\dfrac a4 + \dfrac8a = 3$, so $a^2 - 12a + 32 = 0$, (a − 4)(a − 8) = 0. x = 2 or 3.

Let $5^x = a$: $5a + \dfrac{25}{a} = 126$, so $5a^2 - 126a + 25 = 0$, (5a − 1)(a − 25) = 0, $a = \dfrac15$ or 25. x = −1 or 2.

Let $3^y = a$: (a − 1)(a − 3) = 0, so $3^y = 3^0$ or $3^1$. y = 0 or 1.

Let $4^x = a$: $a^2 - 20a + 64 = 0$, (a − 4)(a − 16) = 0. x = 1 or 2.
x = 1: $5 + \dfrac{125}{5} = 5 + 25 = 30$ ✓. x = 2: $25 + \dfrac{125}{25} = 25 + 5 = 30$ ✓.

Here $pq = 3^{1/3} \times 3^{-1/3} = 1$. Cubing:

$$\begin{aligned} x^3 &= 3 + \frac13 + 3x \\ 3x^3 &= 10 + 9x \\ 3x(x^2 - 3) &= 10 \end{aligned}$$

Proved.

Here pq = −1. Cubing:

$$\begin{aligned} x^3 &= 2 - \frac12 - 3x \\ 2x^3 &= 3 - 6x \\ 2x^3 + 6x - 3 &= 0 \end{aligned}$$

Proved.

Employee A’s: each month is Rs 1,000 more than the one before (a common difference of 1,000). B’s is multiplied by 1.5 each month, a geometric sequence.

A: $\dfrac{5000 + 7000}{2} =$ Rs 6,000 (Jestha’s, the term between).
B: $\sqrt{2000 \times 4500} = \sqrt{9000000} =$ Rs 3,000 (Jestha’s again).

A: $S_5 = \dfrac52(2 \times 5000 + 4 \times 1000) = 35000$.
B: $S_5 = \dfrac{2000(1.5^5 - 1)}{1.5 - 1} = 4000 \times 6.59375 = 26375$.
Difference: 35000 − 26375 = Rs 8,625.

Bishal: $\dfrac62(2a + 5 \times 1000) = 45000$, so 2a + 5000 = 15000, a = Rs 5,000.
Sita: $\dfrac{a(2^6 - 1)}{2 - 1} = 63000$, so 63a = 63000, a = Rs 1,000.

First: 5000 − 1000 = Rs 4,000.
Last: Bishal’s is 5000 + 5 × 1000 = 10000, Sita’s $1000 \times 2^5 = 32000$: a difference of Rs 22,000.

Bishal pays 5,000, 6,000, 7,000, 8,000, …; Sita 1,000, 2,000, 4,000, 8,000, …. The fourth, Rs 8,000 each.

$t_{10} = 2 \times 2^9 =$ 1,024 birds.

$S_{10} = \dfrac{2(2^{10} - 1)}{2 - 1} =$ 2,046 birds.

$\dfrac{a(2^7 - 1)}{2 - 1} = 635$, so 127a = 635: Rs 5.

$5 \times 2^6 =$ Rs 320.

$t_{16} = 500 + 15 \times 500 =$ Rs 8,000.

$S_{16} = \dfrac{16}{2}(500 + 8000) =$ Rs 68,000.

$S_n = \dfrac n2(1000 + 500(n - 1)) = 250n(n + 1)$.
After 19 birthdays: 250 × 19 × 20 = 95000, not yet. After 20: 250 × 20 × 21 = 105000. On his 20th birthday.

First: x(x + 2) = 360, $x^2 + 2x - 360 = 0$, (x + 20)(x − 18) = 0, x = 18: 20 m by 18 m.
Second: y(y + 9) = 360, $y^2 + 9y - 360 = 0$, (y + 24)(y − 15) = 0, y = 15: 24 m by 15 m.

Perimeters: 2(20 + 18) = 76 m and 2(24 + 15) = 78 m. No: the same area doesn’t mean the same perimeter.

The second, by 2 × 10 = Rs 20 (Rs 780 against Rs 760).

Ram (30 − x) years, Sita (25 − x) years.

(30 − x)(25 − x) = 500, $x^2 - 55x + 250 = 0$, (x − 5)(x − 50) = 0. They weren’t born 50 years ago, so x = 5. Check: 25 × 20 = 500 ✓.

(30 + y) + (25 + y) = 99, 2y = 44: in 22 years.

$S_5 = \dfrac52(4 + 4d) = 10 + 10d$ and $S_{10} = 5(4 + 9d) = 20 + 45d$.
The next five: $S_{10} - S_5 = 10 + 35d = 4(10 + 10d)$, so −5d = 30: d = −6.

$t_{20} = 2 + 19 \times (-6) = 2 - 114 = -112$. Shown.

S₅ = 10 + 10 × (−6) = −50 (the terms 2, −4, −10, −16, −22).

The whole is (16 + 2x) m by (12 + 2x) m: (16 + 2x)(12 + 2x) = 320.

$4x^2 + 56x + 192 = 320$, so $x^2 + 14x - 32 = 0$, (x + 16)(x − 2) = 0. A width can’t be negative: 2 m. Check: 20 × 16 = 320 ✓.

By Pythagoras’ theorem: $x^2 + (x + 40)^2 = (x + 80)^2$.

$x^2 + x^2 + 80x + 1600 = x^2 + 160x + 6400$, so $x^2 - 80x - 4800 = 0$, (x − 120)(x + 40) = 0, x = 120.
120 m, 160 m and 200 m. Check: $120^2 + 160^2 = 40000 = 200^2$ ✓.

Perimeter 2(120 + 160) = 560 m; four rounds 2,240 m; 2240 × 15 = Rs 33,600.

$\dfrac{160 \times 120}{20 \times 15} = \dfrac{19200}{300} =$ 64 (8 along the 160 m side and 8 along the 120 m side, so they really fit).

Ramesh (30 − x) years, Sita (25 − x) years.

(30 − x)(25 − x) = 644, $x^2 - 55x + 106 = 0$, (x − 2)(x − 53) = 0. 53 years ago they weren’t born: x = 2. Check: 28 × 23 = 644 ✓.

(30 + y)(25 + y) = 864, $y^2 + 55y - 114 = 0$, (y + 57)(y − 2) = 0: in 2 years (32 × 27 = 864 ✓).

If the westbound car has gone x miles, by Pythagoras’ theorem: $24^2 + x^2 = (3x + 4)^2$.

$576 + x^2 = 9x^2 + 24x + 16$, so $8x^2 + 24x - 560 = 0$, $x^2 + 3x - 70 = 0$, (x + 10)(x − 7) = 0: 7 miles.

3 × 7 + 4 = 25 miles. Check: $24^2 + 7^2 = 625 = 25^2$ ✓.

Time = distance ÷ speed, and 30 minutes is half an hour: $\dfrac{90}{x} - \dfrac{90}{x + 15} = \dfrac12$.

$$\frac{90(x + 15) - 90x}{x(x + 15)} = \frac12$$

$1350 \times 2 = x^2 + 15x$, so $x^2 + 15x - 2700 = 0$, (x + 60)(x − 45) = 0: 45 km/h.

90 ÷ 45 = 2 hours (and at 60 km/h it would have taken $1\dfrac12$ hours ✓).

$$\frac{(a + b) - 2b}{(a - b)(a + b)} = \frac{a - b}{(a - b)(a + b)} = \frac{1}{a + b}$$

$\dfrac{1}{a + b}$

$a^4 + 16a^2 + 256 = (a^2 + 4a + 16)(a^2 - 4a + 16)$. Add the first two fractions first.

The first two: the numerator is $(a - 4)(a^2 + 4a + 16) + (a + 4)(a^2 - 4a + 16) = (a^3 - 64) + (a^3 + 64) = 2a^3$, so

$$\frac{2a^3}{a^4 + 16a^2 + 256} + \frac{128}{a^4 + 16a^2 + 256} = \frac{2(a^3 + 64)}{a^4 + 16a^2 + 256}$$

$a^3 + 64 = (a + 4)(a^2 - 4a + 16)$, which cancels:

$\dfrac{2(a + 4)}{a^2 + 4a + 16}$

The denominators are (a − 4)(a − 5), (a − 3)(a − 4) and (a − 3)(a − 5); the LCM is (a − 3)(a − 4)(a − 5).

$$\frac{(2a - 6)(a - 3) - (a - 1)(a - 5) - (a - 2)(a - 4)}{(a - 3)(a - 4)(a - 5)}$$

The top: $(2a^2 - 12a + 18) - (a^2 - 6a + 5) - (a^2 - 6a + 8) = 5$.

$\dfrac{5}{(a - 3)(a - 4)(a - 5)}$

Over the LCM 2abc, the top is

$$c(a + b)(a + b - c) + a(b + c)(b + c - a) + b(c + a)(c + a - b)$$

Multiplied out, every term with three different letters adds to 6abc and the rest cancel in pairs (for example $a^2c$ from the first and $-a^2c$ from the third). So the sum is $\dfrac{6abc}{2abc}$: 3. Check with a = b = c = 1: 1 + 1 + 1 = 3 ✓.

Let $3^x = a$: $9a + \dfrac9a = 82$, so $9a^2 - 82a + 9 = 0$, (a − 9)(9a − 1) = 0. x = ±2.

$3^x + \dfrac{1}{3^x} = 9 + \dfrac19$: let $3^x = a$, $9a^2 - 82a + 9 = 0$, a = 9 or $\dfrac19$. x = ±2.

Move the 1 across first: $x - 1 = 2^{1/3} + 2^{2/3}$, and $2^{1/3} \times 2^{2/3} = 2$.

Cube $x - 1 = 2^{1/3} + 2^{2/3}$:

$$\begin{aligned} (x - 1)^3 &= 2 + 4 + 3 \times 2 \times (x - 1) \\ x^3 - 3x^2 + 3x - 1 &= 6x \\ x^3 - 3x^2 - 3x &= 1 \end{aligned}$$

So $x(x^2 - 3x - 3) = 1$. Proved.

Cube $x - 3 = 3^{1/3} + 3^{2/3}$, where $3^{1/3} \times 3^{2/3} = 3$:

$$\begin{aligned} (x - 3)^3 &= 3 + 9 + 9(x - 3) \\ x^3 - 9x^2 + 27x - 27 &= 9x - 15 \\ x^3 - 9x^2 + 18x &= 12 \end{aligned}$$

So $x(x^2 - 9x + 18) = 12$. Proved.

Cube $x - 2 = 2^{2/3} - 2^{1/3}$, where $2^{2/3} \times (-2^{1/3}) = -2$:

$$\begin{aligned} (x - 2)^3 &= 4 - 2 - 6(x - 2) \\ x^3 - 6x^2 + 12x - 8 &= 14 - 6x \\ x^3 - 6x^2 + 18x &= 22 \end{aligned}$$

So $x(x^2 - 6x + 18) = 22$. Proved.