Taking logarithms gives \[\ln y=\ln k+a\sqrt x.\] The gradient of the graph is \(a\): \[a=\frac{4.87-2.83}{3.2-1.5}=\frac{2.04}{1.7}=\boxed{1.2}.\]

Using the first point, \[\ln k=2.83-1.2(1.5)=1.03,\qquad k=e^{1.03}=2.80106\ldots=\boxed{2.8}.\]

Take natural logarithms and use the power law: \[(x+y)\ln3=(4x+1)\ln2.\] Rearrange: \[y=\frac{4\ln2-\ln3}{\ln3}x+\frac{\ln2}{\ln3}.\] Since \(4\ln2-\ln3=\ln(16/3)\), \[\boxed{m=\frac{\ln(16/3)}{\ln3}},\qquad\boxed{c=\frac{\ln2}{\ln3}}.\]

Taking natural logarithms gives \(\ln y=\ln k+(x+2)\ln a\). Hence the gradient of the graph is \(\ln a\).

\[\ln a=\frac{4.69-3.22}{4.0-1.5}=0.588,\qquad a=e^{0.588}=1.80038\ldots.\]

Using the first point, \(3.22=\ln k+3.5(0.588)\), so \(\ln k=1.162\) and \(k=e^{1.162}=3.19632\ldots\).

\[\boxed{k=3.2,\quad a=1.8}\quad\text{(2 s.f.)}.\]

Let \(u=e^{2x}>0\). Then \[u^2-8u-48=0,\qquad(u-12)(u+4)=0.\] Reject \(u=-4\), since an exponential is positive. Thus \(e^{2x}=12\) and \[x=\frac12\ln12\approx1.24.\]

The logarithms require \(x>2\). Using the quotient law, \[\ln\left(\frac{3x+5}{x-2}\right)=4\Rightarrow3x+5=e^4(x-2).\] Thus \[x=\frac{2e^4+5}{e^4-3}.\] This is greater than 2, since \(x-2=11/(e^4-3)>0\), so it is valid.

(a) Since \(\ln4>0\), \[x\ln4<\ln0.05\Rightarrow x<\frac{\ln0.05}{\ln4}=-2.16\quad(3\text{ s.f.}).\]

(b) \[-9<3x+8<9\Rightarrow-17<3x<1\Rightarrow-\frac{17}{3}<x<\frac13.\]

(c) Combining these intervals, the integers are \(-5,-4,-3\).

The domain requires \(x>5\). The quotient law gives \[\ln\left(\frac{3x+1}{x-5}\right)=\ln7\Rightarrow3x+1=7(x-5).\] Thus \(4x=36\) and \(x=9\), which is in the domain.

(a) Taking natural logarithms gives \(2y\ln a=3x+k\), so \[y=\frac3{2\ln a}x+\frac{k}{2\ln a}.\] The gradient is \(3/(2\ln a)\).

(b) \[\frac3{2\ln a}=\frac{3.80-0.95}{3.3-0.4}=\frac{57}{58}\Rightarrow\ln a=\frac{29}{19},\quad a=e^{29/19}=4.60.\] At (0.4,0.95), \(k=2(0.95)(29/19)-3(0.4)=1.70\).

Taking logarithms, \(8y\ln5=7x\ln6\), so \[y=\frac{7\ln6}{8\ln5}x,\quad k=0.974.\]

Taking natural logarithms, \[(2x-1)\ln6=\ln5+3x+2.\] Thus \[x=\frac{\ln6+\ln5+2}{2\ln6-3}=9.256\] to 4 significant figures.

\[(4x+3)\ln3=(2x+7)\ln5\Rightarrow x=\frac{7\ln5-3\ln3}{4\ln3-2\ln5}=6.78.\]

Since \(\theta\) is acute, \(\cos\theta=\sqrt{1-4/9}=\sqrt5/3\). Use the addition formula: \[\sin(\theta+60^\circ)=\sin\theta\cos60^\circ+\cos\theta\sin60^\circ=\frac23\frac12+\frac{\sqrt5}{3}\frac{\sqrt3}{2}=\frac13+\frac{\sqrt{15}}6.\]

(a) Where the original expression is defined, \[\frac{3(2\sin\theta\cos\theta)+4\sin^3\theta}{\sin\theta}=6\cos\theta+4\sin^2\theta=4+6\cos\theta-4\cos^2\theta.\]

(b) By (a), \(7+6\cos\theta-4\cos^2\theta=0\). Let \(u=\cos\theta\). Then \[4u^2-6u-7=0,\qquad u=\frac{3\pm\sqrt{37}}4.\] The positive-sign value exceeds 1, so \(u=(3-\sqrt{37})/4\). The unique angle in the stated interval is \[\theta=-\arccos\!\left(\frac{3-\sqrt{37}}4\right)\approx-2.45.\]

(c) Using \(\cos^2\theta=(1+\cos2\theta)/2\), the integrand becomes \(2+6\cos\theta-2\cos2\theta\). Hence \[\int\cdots\,d\theta=2\theta+6\sin\theta-\sin2\theta+C.\]

On the domain where \(\sec\theta\) exists, multiply by \(\cos\theta\) and expand: \[\cos\theta\cos60^\circ+\sin\theta\sin60^\circ=4\cos\theta.\] Thus \[\frac12+\frac{\sqrt3}{2}\tan\theta=4,\qquad\tan\theta=\frac7{\sqrt3}.\] The angles in the given interval are \(\theta=76.1^\circ\) and \(\theta=-103.9^\circ\), to 1 decimal place.

(a) Where the original expression is defined, \[\sin2x(\cot x+3\tan x)=2\sin x\cos x\left(\frac{\cos x}{\sin x}+3\frac{\sin x}{\cos x}\right)\] \[=2\cos^2x+6\sin^2x=(1+\cos2x)+3(1-\cos2x)=4-2\cos2x.\]

(b) Set \(x=\pi/12\), so \(\sin2x=1/2\) and \(\cos2x=\sqrt3/2\). Then \[\cot(\pi/12)+3\tan(\pi/12)=\frac{4-\sqrt3}{1/2}=8-2\sqrt3.\]

(c) \(dy/dx=4\sin2x\). The illustrated first rising-branch point with gradient 4 has \(a=\pi/4\). The illustrated next minimum has \(b=\pi\). Hence \[\int_{\pi/4}^{\pi}(4-2\cos2x)\,dx=[4x-\sin2x]_{\pi/4}^{\pi}=4\pi-(\pi-1)=3\pi+1.\]

(a) Expanding gives \(R\cos\alpha=7\) and \(R\sin\alpha=24\). Thus \(R=\sqrt{7^2+24^2}=25\) and \(\alpha=\tan^{-1}(24/7)=73.74^\circ\).

(b) \(\cos(\theta-\alpha)=18/25\). Hence \(\theta=\alpha\pm\cos^{-1}(18/25)\), giving \(29.8^\circ\) and \(117.7^\circ\) in the required interval.

(c) The denominator is \(50+25\cos(\beta/2-\alpha)\), with minimum \(25\). Therefore \(V_{\max}=150/25=6\). This requires \(\beta/2-\alpha=180^\circ+360^\circ n\). The smallest positive \(\beta\) is \(2(\alpha+180^\circ)=507.5^\circ\), using the unrounded \(\alpha\).

Use \(\tan^2\theta=\sec^2\theta-1\): \[6\sec^2\theta-17\sec\theta-14=0,\qquad(2\sec\theta-7)(3\sec\theta+2)=0.\] Hence \(\sec\theta=7/2\) or \(-2/3\). The second is impossible for a real angle because \(|\sec\theta|\ge1\). Thus \(\cos\theta=2/7\), giving \(\theta=73.4^\circ,286.6^\circ\).

(a) Expand both brackets: \[4(\tfrac12\sin\theta+\tfrac{\sqrt3}{2}\cos\theta)(\tfrac12\cos\theta+\tfrac{\sqrt3}{2}\sin\theta)=4\sin\theta\cos\theta+\sqrt3(\sin^2\theta+\cos^2\theta)=2\sin2\theta+\sqrt3.\] Equivalently, using \(2\sin A\cos B=\sin(A+B)+\sin(A-B)\), \[4\sin(\theta+\pi/3)\cos(\theta-\pi/3)=2\sin2\theta+2\sin(2\pi/3)=2\sin2\theta+\sqrt3.\]

(b) Put \(\theta=3\pi/8\). Then the angles are \(17\pi/24\) and \(\pi/24\), and the value is \(\sqrt3+2\sin(3\pi/4)=\sqrt3+\sqrt2\).

(c) Apply the identity with \(\theta=2x\). Then \[\int_0^{\pi/8}(\sqrt3+2\sin4x)\,dx=\left[\sqrt3x-\frac12\cos4x\right]_0^{\pi/8}=\frac{\sqrt3\pi}{8}+\frac12.\]

Let \(u=\tan\theta\). In the original domain \(u\ne0\) and \(1+\sqrt3u\ne0\). The tangent subtraction formula gives \[\frac{u-\sqrt3}{1+\sqrt3u}=\frac3u,\qquad u^2-4\sqrt3u-3=0.\] Hence \(u=2\sqrt3\pm\sqrt{15}\). Both roots satisfy the original domain restrictions, and \(\theta=\tan^{-1}u\) gives \(-22.2^\circ,82.2^\circ\).

Taking natural logarithms gives \[\ln14-2x=(x+1)\ln5.\] Collecting terms, \[x(2+\ln5)=\ln14-\ln5=\ln(14/5),\] so \[x=\frac{\ln(14/5)}{2+\ln5}\approx0.285.\]

Taking logarithms, \[\ln y=\ln A+k\ln x.\] Therefore \(k\) is the gradient: \[k=\frac{3.47-2.87}{0.81-0.56}=2.4.\] Substituting \((0.56,2.87)\), \[\ln A=2.87-2.4(0.56)=1.526.\] Hence \[A=e^{1.526}\approx4.6\quad\text{(2 significant figures)}.\]

(a) Taking logarithms, \[\ln y=(2x-a)\ln4=(2\ln4)x-a\ln4.\] The gradient is \(2\ln4=\ln(4^2)=\ln16\).

(b) The vertical intercept is \(-a\ln4\), so \[a\approx\frac{20.8}{\ln4}\approx15.004.\] Since \(a\) is an integer, \(a=15\). Indeed, \(-15\ln4\approx-20.7944\), which rounds to \(-20.8\) to 3 significant figures.

(a) Taking logarithms, \[\ln y=2a\ln3+x\ln a.\] The gradient is \(\ln a=0.239\), so \[a=e^{0.239}\approx1.27\quad\text{(3 significant figures)}.\]

(b) Retain the unrounded value \(a=e^{0.239}\). Substituting \(y=36\), \[\ln36=2e^{0.239}\ln3+0.239x.\] Therefore \[x=\frac{\ln36-2e^{0.239}\ln3}{0.239}\approx\boxed{3.32}.\]

Taking natural logarithms gives \[y\ln a=\ln k+\ln x,\qquad y=\frac{\ln x}{\ln a}+\frac{\ln k}{\ln a}.\] Thus the gradient is \[\frac1{\ln a}=\frac{9.00-6.36}{2.58-1.03}=\frac{2.64}{1.55}.\] Hence \[\ln a=\frac{1.55}{2.64},\qquad a=e^{1.55/2.64}=1.79880\ldots.\] Substitute the first point: \[\ln k=6.36\left(\frac{1.55}{2.64}\right)-1.03=2.70409\ldots,\qquad k=e^{2.70409\ldots}=14.9407\ldots.\] Therefore, to 2 significant figures, \[\boxed{a=1.8,\qquad k=15}.\]

(a) The logarithms require \(x>0\). Combine and exponentiate: \[\ln\frac{x+2}{x}=\ln9,\qquad x+2=9x,\qquad\boxed{x=1/4}.\] This satisfies the original domain.

(b) Replace x by \(\cot y\): \(\cot y=1/4\), so \(\tan y=4\). The stated first-quadrant interval contains exactly one solution: \[y=\arctan4=1.325817\ldots,\qquad\boxed{y=1.326\text{ radians}}.\]