Sec 4 A Math: Surds, Indices & Logarithms, practice questions & worked solutions

Perpendicular height $h$ $$\begin{aligned} &= \frac{2\sqrt{6} + 3}{\sqrt{6} - 1}\\[2pt] &= \frac{2\sqrt{6} + 3}{\sqrt{6} - 1} \times \frac{\sqrt{6} + 1}{\sqrt{6} + 1}\\[2pt] &= \frac{12 + 3 + 5\sqrt{6}}{5}\\[2pt] &= \left(3 + \sqrt{6}\right)\ \text{cm} \end{aligned}$$

$$\begin{aligned} 2^{\frac{3x}{2}} &= 2^{x - 2}\\ \frac{3x}{2} &= x - 2\\ x &= -4\\ \sqrt{8^{-4}} &= \frac{1}{64} \end{aligned}$$

(a)

$$\begin{aligned} \text{Area of square} &= 11 + \sqrt{120}\\ \left(\sqrt{a} + \sqrt{b}\right)^2 &= 11 + 2\sqrt{30}\\ a + b + 2\sqrt{ab} &= 11 + 2\sqrt{30} \end{aligned}$$ Equating the rational part, $$\begin{aligned} a + b &= 11\\ b &= 11 - a \text{------(1)} \end{aligned}$$ Equating the irrational part, $$\begin{aligned} 2\sqrt{ab} &= 2\sqrt{30}\\ ab &= 30 \text{------(2)} \end{aligned}$$ Subst. (1) into (2), $$\begin{aligned} a(11 - a) &= 30\\ 11a - a^2 &= 30\\ a^2 - 11a + 30 &= 0\\ (a - 5)(a - 6) &= 0\\ a = 5 \ &\text{or}\ a = 6 \end{aligned}$$ Corresponding $b$ values: $$b = 6 \ \text{or}\ b = 5$$ Since $a < b$ $$a = 5, b = 6$$

$$\begin{aligned} 29000 &= 38000\,e^{-3k}\\ e^{-3k} &= \frac{29}{38}\\ -3k &= \ln\frac{29}{38}\\ k &= -\frac{1}{3}\ln\frac{29}{38} \approx 0.090096 \end{aligned}$$ $$\begin{aligned} 19000 &= 38000\,e^{-0.090096t}\\ e^{-0.090096t} &= \frac{1}{2}\\ -0.090096t &= \ln\frac{1}{2}\\ t &\approx 7.69\ \text{years} \end{aligned}$$ The year is likely to be $2032$.

$$\begin{aligned} \log_4 x-\frac{3}{\log_x 8} &= \frac{\log_2 x}{\log_2 4}-3\div\!\left(\frac{\log_2 8}{\log_2 x}\right)\\ &= \frac{\log_2 x}{2\log_2 2}-3\div\!\left(\frac{3\log_2 2}{\log_2 x}\right)\\ &= \frac{\log_2 x}{2\log_2 2}-3\times\!\left(\frac{\log_2 x}{3\log_2 2}\right)\\ &= \frac{p}{2}-p\\ &= -\frac{p}{2} \end{aligned}$$ $$\begin{aligned} -\frac{p}{2} &= p^2\\ -p &= 2p^2\\ 2p^2+p &= 0\\ p(2p+1) &= 0\\ p=0,\;&p=-\frac{1}{2} \end{aligned}$$ $$\begin{aligned} \log_2 x &= 0\\ x &= 1\ (\text{rej}) \end{aligned}$$ $$\begin{aligned} \log_2 x &= -\frac{1}{2}\\ x &= \frac{1}{\sqrt{2}} \end{aligned}$$

(a)

At $t = 3$, $$\begin{aligned} 1000 &= 500(2)^{3k}\\ 2 &= (2)^{3k}\\ 3k &= 1\\ k &= \frac{1}{3} \end{aligned}$$

(b)

$$\begin{aligned} N &\leq 3000\\ 500(2)^{\frac{t}{3}} &\leq 3000\\ (2)^{\frac{t}{3}} &\leq 6\\ \frac{t}{3} &\leq \frac{\lg 6}{\lg 2}\\ t &\leq 7.75 \end{aligned}$$ Maximum number of hours that it will still be safe to consume $= 7$ hours

(a)

$$2^{2x+1} = 5(2^x) + 3$$ Let $u = 2^x$. $$2u^2 - 5u - 3 = 0$$

(b)

$$\begin{aligned} (2u + 1)(u - 3) &= 0\\ u = -\tfrac{1}{2} \ &\text{or}\ u = 3 \end{aligned}$$ $2^x > 0$ therefore $2^x \neq -\dfrac{1}{2}$ $$\begin{aligned} 2^x &= 3\\ \lg 2^x &= \lg 3\\ x &= \frac{\lg 3}{\lg 2}\\ &= 1.6 \ (\text{2 sf}) \end{aligned}$$

(a)

$$\begin{aligned} 9^x+6^x &= 6(4^x)\\ \frac{9^x}{4^x}+\frac{6^x}{4^x} &= 6\\ \frac{3^{2x}}{2^{2x}}+\left(\frac{6}{4}\right)^x-6 &= 0\\ \left(\frac{3}{2}\right)^{2x}+\left(\frac{3}{2}\right)^x-6 &= 0 \end{aligned}$$ When $u=\left(\dfrac{3}{2}\right)^x$, $u^2+u-6=0$.

(b)

$$\begin{aligned} (u+3)(u-2) &= 0\\ u=2 \;&\text{or}\; -3\ (\text{n.a.})\\ \left(\frac{3}{2}\right)^x &= 2\\ x = \frac{\ln 2}{\ln\frac{3}{2}} \;&\text{or}\; \frac{\lg 2}{\lg\frac{3}{2}} = 1.71 \end{aligned}$$

(a)

$$\begin{aligned} y &= kx + 6 \text{---- (1)}\\ 2x^2 - xy &= 3 \text{-----(2)} \end{aligned}$$ Sub (1) into (2): $$\begin{aligned} 2x^2 - x(kx + 6) &= 3\\ 2x^2 - kx^2 - 6x &= 3\\ (2 - k)x^2 - 6x - 3 &= 0 \end{aligned}$$ $$\begin{aligned} \text{Discriminant} &= (-6)^2 - 4(2 - k)(-3)\\ &= 60 - 12k \end{aligned}$$ For tangent, $$\begin{aligned} 60 - 12k &= 0\\ k &= 5 \end{aligned}$$

(b)

$$\begin{aligned} \left(4 - 2\sqrt{2}\right)x + \sqrt{2} &= 3x\sqrt{2} - 1\\ 4x - 2x\sqrt{2} + \sqrt{2} &= 3x\sqrt{2} - 1\\ \sqrt{2} + 1 &= 3x\sqrt{2} + 2x\sqrt{2} - 4x\\ \sqrt{2} + 1 &= x\left(5\sqrt{2} - 4\right)\\ x &= \frac{\sqrt{2} + 1}{5\sqrt{2} - 4} \times \frac{5\sqrt{2} + 4}{5\sqrt{2} + 4}\\ &= \frac{5(2) + 4\sqrt{2} + 5\sqrt{2} + 4}{\left(5\sqrt{2}\right)^2 - 4^2}\\ &= \frac{9\sqrt{2} + 14}{50 - 16}\\ &= \frac{9}{34}\sqrt{2} + \frac{7}{17} \end{aligned}$$

(a)

Given that $$\log_7 x = m \text{and} \log_7 y = 2n$$ $$x = 7^m \qquad y = 7^{2n}$$ $$\begin{aligned} xy &= \left(7^m\right)\left(7^{2n}\right)\\ xy &= 7^{m + 2n} \end{aligned}$$

(b)

$$\begin{aligned} \log_3 3x^2 + 4 &= \log_x 27\\ \log_3 3 + 2\log_3 x + 4 &= \log_x 27\\ 2\log_3 x + 5 &= \frac{\log_3 3^3}{\log_3 x}\\ 2\log_3 x + 5 &= \frac{3}{\log_3 x} \end{aligned}$$ Let $y = \log_3 x$, $$\begin{aligned} 2y + 5 &= \frac{3}{y}\\ 2y^2 + 5y - 3 &= 0\\ (2y - 1)(y + 3) &= 0\\ y = \frac{1}{2} \text{or} y &= -3\\ \log_3 x = \frac{1}{2} \text{or} \log_3 x &= -3\\ x = 3^{\frac{1}{2}} \text{or} x &= 3^{-3}\\ = \sqrt{3} \text{or} &= \frac{1}{27} \end{aligned}$$