Sec 3 A Math: Surds, Indices & Logarithms, practice questions & worked solutions

$$\begin{aligned} \text{Height} &= \frac{9+5\sqrt{5}}{4+2\sqrt{5}}\cdot\frac{4-2\sqrt{5}}{4-2\sqrt{5}}\\ &= \frac{36-18\sqrt{5}+20\sqrt{5}-50}{16-20}\\ &= \frac{-14+2\sqrt{5}}{-4}\\ &= \frac{7}{2}-\frac{1}{2}\sqrt{5}\\ &= \frac{1}{2}\!\left(7-\sqrt{5}\right)\ \text{cm} \end{aligned}$$

$$\begin{aligned} 2^{2(x+2)} \times 5 &= 5^{x+3} \times 8^x\\ 2^4(2^{2x}) \times 5 &= 5^3(5^x) \times 2^{3x}\\ \frac{2^{2x}}{(2^{3x})(5^x)} &= \frac{5^3}{2^4 \times 5}\\ \frac{1}{(2^x)(5^x)} &= \frac{25}{16}\\ \frac{1}{10^x} &= \left(\frac{5}{4}\right)^2\\ 10^x &= \left(\frac{4}{5}\right)^2\\ x &= \lg\left(\frac{4}{5}\right)^2\\ x &= 2\lg\left(\frac{4}{5}\right) \end{aligned}$$

Or

$$\begin{aligned} 2^{2(x+2)} \times 5 &= 5^{x+3} \times 8^x\\ 2^{2x+4} \times 5 &= 5^{x+3} \times 2^{3x}\\ \frac{2^{2x+4}}{2^{3x}} &= \frac{5^{x+3}}{5}\\ \frac{2^4}{2^x} &= 5^x 5^2\\ 10^x &= \frac{2^4}{5^2}\\ 10^x &= \frac{4^2}{5^2}\\ x &= \lg\left(\frac{4}{5}\right)^2\\ x &= 2\lg\left(\frac{4}{5}\right) \end{aligned}$$

$$\begin{aligned} 8-\sqrt{11-2x} &= -x\\ 8+x &= \sqrt{11-2x}\\ 64+16x+x^2 &= 11-2x\\ x^2+18x+53 &= 0\\ x &= \frac{-18\pm\sqrt{18^2-4(1)(53)}}{2(1)}\\ &= \frac{-18\pm\sqrt{112}}{2}\\ &= -9\pm\frac{4\sqrt{7}}{2}\\ &= -9+2\sqrt{7} \text{or} -9-2\sqrt{7}\ \text{(rej.)} \end{aligned}$$

$$\begin{aligned} 3\log_5 y-\log_y 5 &= 2\\ 3\log_5 y-\frac{\log_5 5}{\log_5 y} &= 2\\ 3\log_5 y-\frac{1}{\log_5 y} &= 2\\ 3(\log_5 y)^2-2\log_5 y-1 &= 0\\ (3\log_5 y+1)(\log_5 y-1) &= 0\\ \log_5 y=1 \;&\text{or}\; -\frac{1}{3}\\ y=5 \;\text{or}\; y&=5^{-\frac{1}{3}}=0.585\ (\text{3 s.f.}) \end{aligned}$$

At $t = 4$ hour, $N = 2N_0$ $$\begin{aligned} 2N_0 &= N_0 e^{4k}\\ e^{4k} &= 2\\ 4k &= \ln 2\\ k &= \frac{1}{4}\ln 2 \end{aligned}$$

$$\begin{aligned} 2^{2x-1} &= 2^{x+3} - 24\\ \tfrac{1}{2}\left(2^x\right)^2 &= 2^3 \times 2^x - 24 \end{aligned}$$ Let $u = 2^x$. $$\begin{aligned} \tfrac{1}{2}(u)^2 &= 2^3 \times u - 24\\ \tfrac{1}{2}u^2 &= 8u - 24\\ u^2 &= 16u - 48\\ u^2 - 16u + 48 &= 0\\ (u - 12)(u - 4) &= 0\\ u = 12 \ &\text{or} \ u = 4\\ 2^x = 12 \ &\text{or} \ 2^x = 4\\ \lg 2^x = \lg 12 \ &\text{or} \ 2^x = 2^2\\ x = \frac{\lg 12}{\lg 2} \ &\text{or} \ x = 2\\ x \approx 3.58496&\\ x \approx 3.6& \end{aligned}$$

$$\begin{aligned} x\sqrt{18} &= 3x + \sqrt{32}\\ 3\sqrt{2}\,x - 3x &= 4\sqrt{2}\\ 3x\left(\sqrt{2} - 1\right) &= 4\sqrt{2}\\ 3x &= \frac{4\sqrt{2}}{\sqrt{2} - 1}\\ &= \frac{4\sqrt{2}\,\left(\sqrt{2} + 1\right)}{\left(\sqrt{2} - 1\right)\left(\sqrt{2} + 1\right)}\\ &= \frac{8 + 4\sqrt{2}}{\left(\sqrt{2}\right)^2 - (1)^2}\\ x &= \frac{8 + 4\sqrt{2}}{3} \end{aligned}$$

Or

$$\begin{aligned} x\sqrt{18} &= 3x + \sqrt{32}\\ 3\sqrt{2}\,x &= 3x + 4\sqrt{2}\\ \left(3\sqrt{2} - 3\right)x &= 4\sqrt{2}\\ x &= \frac{4\sqrt{2}}{3\sqrt{2} - 3}\\ &= \frac{4\sqrt{2}\,\left(3\sqrt{2} + 3\right)}{\left(3\sqrt{2} - 3\right)\left(3\sqrt{2} + 3\right)}\\ &= \frac{24 + 12\sqrt{2}}{\left(3\sqrt{2}\right)^2 - (3)^2}\\ &= \frac{24 + 12\sqrt{2}}{9}\\ &= \frac{8 + 4\sqrt{2}}{3} \end{aligned}$$

(a)

When $t = 0$, $N = 7$. $$\begin{aligned} 7 &= 3 + a\mathrm{e}^{-k(0)}\\ a &= 7 - 3\\ &= 4 \end{aligned}$$ When $t = 1$, $N = 5$. $$\begin{aligned} 5 &= 3 + 4\mathrm{e}^{-k(1)}\\ 2 &= 4\mathrm{e}^{-k}\\ \mathrm{e}^{-k} &= \frac{2}{4}\\ -k &= \ln\frac{1}{2}\\ k &= \ln 2 \end{aligned}$$

(b)

$$N = 3 + 4\mathrm{e}^{-(\ln 2)t}$$ When $t$ is large, $\mathrm{e}^{-(\ln 2)t}$ becomes close to zero.

Thus, the approximate size is $3$ million.

(a)

$$\begin{aligned} \frac{6\sqrt{2}-9}{\sqrt{2}-1}\times\frac{\sqrt{2}+1}{\sqrt{2}+1} &= \frac{12+6\sqrt{2}-9\sqrt{2}-9}{2-1}\\ &= 3-3\sqrt{2} \end{aligned}$$

(b)

Method 1

$$\begin{aligned} (3-3\sqrt{2})^2+p(3-3\sqrt{2})+q &= 0\\ 9-18\sqrt{2}+18+3p-3p\sqrt{2}+q &= 0\\ -(18+3p)\sqrt{2}+27+3p+q &= 0\\ 18+3p &= 0 \text{---(1)}\\ 27+3p+q &= 0 \text{---(2)}\\ p &= -6\\ q &= -9 \end{aligned}$$

Method 2

$$\begin{aligned} x &= \frac{-p\pm\sqrt{p^2-4q}}{2}\\ 3-3\sqrt{2} &= \frac{-p\pm\sqrt{p^2-4q}}{2}\\ 6-6\sqrt{2} &= -p\pm\sqrt{p^2-4q} \end{aligned}$$ Comparing, $$\begin{aligned} p &= -6\\ -\sqrt{p^2-4q} &= -6\sqrt{2}\\ p^2-4q &= 36\times 2\\ (-6)^2-4q &= 72\\ q &= -9 \end{aligned}$$

(a)

$$\begin{aligned} \log_2(x + 2) - \log_{\sqrt{2}}(x - 1) &= 1\\ \log_2(x + 2) - \frac{\log_2(x - 1)}{\log_2\sqrt{2}} &= 1\\ \log_2(x + 2) - \frac{\log_2(x - 1)}{\log_2 2^{\frac{1}{2}}} &= 1\\ \log_2(x + 2) - \frac{\log_2(x - 1)}{\frac{1}{2}\log_2 2} &= 1\\ \log_2(x + 2) - \frac{\log_2(x - 1)}{\frac{1}{2}} &= 1\\ \frac{1}{2}\log_2(x + 2) - \log_2(x - 1) &= \frac{1}{2}\\ \log_2(x + 2) - 2\log_2(x - 1) &= 1\\ \log_2\left(\frac{x + 2}{(x - 1)^2}\right) &= 1\\ \frac{x + 2}{(x - 1)^2} &= 2\\ x + 2 &= 2\left(x^2 - 2x + 1\right)\\ x + 2 &= 2x^2 - 4x + 2\\ 2x^2 - 4x + 2 - x - 2 &= 0\\ 2x^2 - 5x &= 0\\ x(2x - 5) &= 0\\ x = 0 \ &\text{or} \ 2x - 5 = 0\\ x = 0 \ (\text{N.A}) \ &\text{or} \ x = 2\tfrac{1}{2} \end{aligned}$$

(b)

(i) $$\begin{aligned} \lg a - \lg b &= \lg(a + b)\\ \lg\frac{a}{b} &= \lg(a + b)\\ \frac{a}{b} &= (a + b)\\ a &= ab + b^2\\ a - ab &= b^2\\ a &= \frac{b^2}{1 - b} \end{aligned}$$

(ii) $$\begin{aligned} \text{Since } a > 0, \ &\therefore \ \frac{b^2}{1 - b} > 0\\ \because \ b^2 > 0, \ &\therefore \ 1 - b > 0\\ &b < 1\\ &\therefore\ 0 < b < 1 \end{aligned}$$