Problems

Solution:

1. \(\,\) \begin{align*} && 0 &= nx^2 + 2\sqrt{3n^2+50}x + 2n + 15 \\ && 0 &> \Delta = 4(3n^2+5) - 4\cdot n \cdot (2n + 15) \\ \Leftrightarrow && 0 &> n^2-15n+5\\ \text{cv}: && n &= \frac{15 \pm \sqrt{225 - 20}}{2} \\ &&&\approx \frac{15\pm14.x}{2}\\ \Leftrightarrow &&n &\in \{1, 2, \cdots, 14\} \end{align*}
2. \(\,\) \begin{align*} && 0 &> \Delta = 4(pn^2+q) - 4\cdot n \cdot (rn+s) \\ \Leftrightarrow && 0&>(p-r)n^2-sn+q \end{align*} Which is always true if \(r > p\) and \(s^2 < 4q(p-r)\)
3. \(\,\) \begin{align*} && 0 &= x^2 + 2\sqrt{p+q}x+ r+s \\ && 0 &\leq \Delta = 4(p+q) - 4(r+s) \\ && 0 &\leq 1 + s^2/8 - s \\ \text{c.v}: && s &= \frac{1 \pm \sqrt{1-4 \cdot \frac{1}{8}}}{2\cdot\frac18} \\ &&&= 4 \pm 4\sqrt{\frac12} \\ &&&= 4 \pm 2\sqrt{2} \\ \Rightarrow && s \leq 4 - 2\sqrt{2} &\text{ or } s \geq 4 + 2\sqrt{2} \end{align*}

Solution: \begin{align*} && S_2(x) &= e^{x^3} \frac{d^2}{\d x^2} \left [e^{-x^3} \right] \\ &&&= e^{x^3} \frac{d}{\d x} \left [e^{-x^3}(-3x^2) \right] \\ &&&= e^{x^3} \left [e^{-x^3}(9x^4-6x) \right] \\ &&&=9x^4-6x \\ \\ && S_3(x) &= e^{x^3} \frac{\d^3}{\d x^3} \left [ e^{-x^3} \right]\\ &&&= e^{x^3} \frac{\d}{\d x} \left [ e^{-x^3}(9x^4-6x) \right ] \\ &&&= e^{x^3} e^{-x^3}\left [ (-3x^2)(9x^4-6x)+(36x^3-6) \right ] \\ &&&= -27x^6 +54x^3-6 \end{align*} Claim: \(S_n\) is a polynomial of degree \(2n\) with leading coefficient \((-3)^n\). Proof: Clearly this is true for \(n = 1, 2, 3\) by demonstration. Suppose it is true for some \(n = k\), then \begin{align*} && S_k(x) &= e^{x^2} \frac{\d^k}{\d x^k} \left [ e^{x^3}\right] \\ && (-3)^kx^{2k} +\cdots &= e^{x^3} \frac{\d^k}{\d x^k} \left [ e^{x^3}\right] \\ \Rightarrow && \frac{\d^k}{\d x^k} \left [ e^{x^3}\right] &= e^{-x^3} \left ( (-3)^kx^{2k} +\cdots\right) \\ \Rightarrow && \frac{\d^k}{\d x^k}\left [ e^{x^3}\right] &= e^{-x^3} (-3x^2)\left ( (-3)^kx^{2k} +\cdots\right) + e^{-x^3} S_k'(x) \\ &&&= e^{-x^3} \left (\underbrace{ (-3)^{k+1}x^{2k+2} + \cdots + S_k'(x)}_{\deg =2k+2}\right) \\ \Rightarrow && S_{k+1}(x) &= (-3)^{k+1}x^{2k+2} + \cdots + S_k'(x) \end{align*} And therefore \(S_{k+1}\) is a polynomial degree \(2(k+1)\) with leading coefficient \((-3)^{k+1}\) so by induction it's true for all \(n\). If \(S'_n(a) = 0\) then \(S_{n+1}(a) = (-3a^2)S_n(a) + S_n'(a) \Rightarrow S_{n+1}(a)S_n(a) = -3 (aS_n(a))^2 \leq 0\)

Solution: \begin{align*} && 5 \cos x + 12\sin x &= 7 \\ \Rightarrow && 5 \cos x - 7 &= -12 \sin x \\ \Rightarrow && 25 \cos^2 x - 70\cos x + 49 &= 144 \sin^2 x \\ \Rightarrow && 25 \cos^2 x - 70\cos x + 49 &= 144 (1-\cos^2 x) \\ \Rightarrow && 169 \cos^2 x - 70 \cos x -95 &= 0 \\ \Rightarrow && \cos \alpha &= \frac{70 \pm \sqrt{70^2 - 4 \cdot 169 \cdot (-95)}}{2 \cdot 169} \\ &&&= \frac{35 \pm \sqrt{35^2 + 169 \cdot 95} }{169} \\ &&&= \frac{35 \pm 12\sqrt{120}}{169} \end{align*} If \(\frac12 \pi < \alpha < \pi\) then \(\cos \alpha\) is negative, in particular \(\cos \alpha = \frac{35 -12\sqrt{120}}{169}\). Since \(\cos\) is decreasing over this range, if \(\cos \alpha > \cos \frac34 \pi = -\frac{\sqrt{2}}2\), then we will have shown \(\alpha < \frac34 \pi\) \begin{align*} && \cos \alpha &= \frac{35 - 12 \sqrt{120}}{169} \\ &&&> \frac{35 - 12 \cdot \sqrt{121}}{169} \\ &&&= \frac{35 - 12 \cdot 11}{169} \\ &&&= \frac{35 - 132}{169} \\ &&&= -\frac{97}{169} \\ &&&> -\frac{8}{13} \end{align*} but \(\left ( \frac{8}{13} \right)^2 = \frac{64}{169} < \frac12\), so we are done.

Solution: \begin{align*} && 0 &= V\sin \alpha t-\frac12 gt^2 \\ \Rightarrow && t &= \frac{2V \sin \alpha}{g} \\ && R &= V \cos \alpha \, t \\ &&&= \frac{2V^2 \sin \alpha \cos \alpha}{g} \\ &&&= \frac{V^2 \sin 2 \alpha}{g} \end{align*} Therefore the max distance is \(\frac{V^2}{g}\), therefore we cannot hit the target if \(R > \frac{V^2}{g} \Rightarrow gR > V^2\). We have \(\beta = \omega t \Rightarrow t = \frac{\beta}{\omega}\) \begin{align*} && \sin \alpha &= \frac{gt}{2V} \\ && \cos \alpha &= \frac{R}{Vt} \\ \Rightarrow && 1 &= \left (\frac{gt}{2V} \right)^2 + \left ( \frac{R}{Vt} \right)^2 \\ &&&= \left (\frac{g\beta}{2V \omega} \right)^2 + \left ( \frac{R\omega}{V\beta} \right)^2 \\ &&&= \frac{g^2 \beta^2}{4 V^2 \omega^2} + \frac{R^2 \omega^2}{V^2 \beta ^2} \\ \Rightarrow && 4V^2 \omega^2 \beta^2 &= g^2 \beta^4 + 4R^2 \omega^4 \\ \Rightarrow && 0 &= g^2 \beta^4 - 4\omega^2 V^2 \beta^2+4R^2\omega^4 \end{align*} This (quadratic) equation in terms of \(\beta^2\) has two solution if \(\Delta = 16\omega^4V^4-16g^2R^2\omega^4 =16\omega^4(V^4-g^2R^2) > 0\) since \(V^2 > gR\). Since \(\beta > 0\) there are exactly two solutions, once we have values for \(\beta\). First notice, \begin{align*} && \beta_1^2 + \beta_2^2 &= \frac{4\omega^2V^2}{g^2} \\ && \beta_1^2\beta_2^2 &= \frac{4R^2\omega^4}{g^2} \end{align*} Then notice the positions of \(P_1\) and \(P_2\) are \((R\cos \beta_1 , R\sin \beta_1)\) and \((R\cos \beta_2, R\sin \beta_2)\). \begin{align*} && d^2 &= R^2\left ( \cos \beta_1 - \cos \beta_2 \right)^2 + R^2 \left ( \sin \beta_1 - \sin \beta_2 \right)^2 \\ &&&= 2R^2 - 2R^2(\cos \beta_1 \cos \beta_2 + \sin \beta_1 \sin \beta_2) \\ &&&= 2R^2-2R^2\cos(\beta_1 - \beta_2) \\ &&&= 2R^2 \left (1-\cos(\sqrt{(\beta_1-\beta_2)^2} \right ) \\ &&&= 2R^2 \left (1 - \cos\left ( \sqrt{\frac{4\omega^2 V^2}{g^2} - \frac{4R\omega^2}{g}} \right) \right) \\ &&&= 2R^2 \left (1 - \cos\left (\frac{2\omega}{g} \sqrt{V^2 - Rg} \right) \right) \\ &&&= 4 R^2 \sin^2 \left (\frac{\omega}{g} \sqrt{V^2 - Rg} \right) \end{align*} which gives us the required result.

Solution:

1. \(\,\) \begin{align*} && f(x) &= (1+x^2)e^x \\ && f'(x) &= 2xe^x + (1+x^2)e^x \\ &&&= (1+x)^2e^x \geq 0 \end{align*}
   

   + - ↺
   As \(x \to -\infty\), \(f(x) \to 0\) and as \(x \to \infty\), \(f(x) \to +\infty\) and since \(f\) is strictly increasing we have exactly one solution to \(f(x) = k\) on \((0,\infty)\). Since \(f(x) > 0\) there are no solutions if \(k \leq 0\).
2. Considering the function \(g(x) = (e^x-1)-k\tan^{-1} x \) then \(g'(x) = e^x - \frac{k}{1+x^2}\) therefore \(g'(x)\) has exactly one turning point when \(k > 0\) and \(0\) otherwise at the root of \(f(x) = k\) Notice also that \(g(0) = 0\) so we already have one solution, and \(g'(0) = 1 - k > 0\). Notice from our sketch that if \(0 < k < 1\) the root for \(f(x) = k\) has \(x \leq 0\), so our turning point is to the left of the origin and we are interested in the behaviour of \(g(x)\) as \(x \to -\infty\). (ie do we cross the axis again). \(\lim_{x \to -\infty} \left [ e^x - 1 - k \tan^{-1} x \right] = 0 - 1 +k \frac{\pi}{2} = k \frac{\pi}{2} - 1\). if this is positive, ie if \(k > \frac{2}{\pi}\) there are two solutions, otherwise there is only one real root.

Solution: Note that each of the \(F\) faces have \(m\) edges. If we count edges from each face we will get \(mF\) edges, but this counts each edge twice, so \(E = \frac{m}{2}F\). Similarly each edge goes into \(2\) vertices, but this counts each vertex \(n\) times, therefore \(V = \frac{2E}{n} = \frac{m}{n}F\) Therefore \begin{align*} && 2 &= V - E + F \\ &&&= \frac{m}{n}F - \frac{m}{2}F + F \\ &&&= F \left ( \frac{2m-nm+2n}{2n} \right) \\ &&&= F \left ( \frac{4-(n-2)(m-2)}{2n} \right) \\ \Rightarrow && F &= \frac{4n}{4-(n-2)(m-2)} = \frac{4n}{h} \end{align*} Notice that \(1 \leq h \leq 4\), so If \(h = 4\), then \(n = 2\) or \(m = 2\) but we can't have two-sided polygons or polyhedra with two faces making a vertex so this is not possible. If \(h = 3\) then \((n-2)(m-2) = 1\) and \(3 \mid n\), so we have \(n = 3, m = 3\). ie we have triangular faces meeting at \(3\) per point. We also have \(F = \frac{4 \cdot 3 }3\) which is \(4\) faces so this is a tetrahedron and \((V,E, F) = (4, 6, 4)\) If \(h = 2\) then \((n-2)(m-2) = 2\): Case 1: \(n = 4, m = 3\) which is four triangles meeting at each vertex with \((V,E,F) = (6, 12, 8)\), ie an octohedron. Case 2: \(n = 3, m = 4\) so we have three squares meeting at each vertex, with \((V,E,F) = (8,12,6)\) which is a cube. If \(h = 1\) then \((n-2)(m-2) = 3\) Case 1: \(n = 5, m = 3\) which is five triangles meeting at each vertex, with \((V,E,F) = ( 12,30, 20)\) ie an icosahedron. Case 2: \(n = 3, m = 5\) which is three pentagons meeting at each vertex, with \((V,E,F) = (20, 30,12)\) which is a dodecahedron. These are all the possible cases and hence we have found the five platonic solids.

Solution:

1. True. \begin{align*} && \ln a \cdot \ln b &= \ln b \cdot \ln a \\ \Leftrightarrow && \exp ( \ln a \cdot \ln b) &= \exp ( \ln b \cdot \ln a) \\ \Leftrightarrow && \exp ( \ln a )^{\ln b} &= \exp ( \ln b )^{\ln a} \\ \Leftrightarrow && a^{\ln b} &= b^{\ln a} \\ \end{align*}
2. False. Consider \(\theta = 0\). We'd need \(\cos 0 = 1 = \sin 1\), but \(0 < 1 < \frac{\pi}{2}\) so \(\sin 1 \neq 1\)
3. False. If the polynomial has positive degree, then as \(n \to \infty\), \(\P(x) \to \pm \infty\), in particular it must be well outside the interval \([-1,1]\). Therefore it can't be within \(10^{-6}\) of \(\cos \theta\) which is confined to that interval. The only polynomial which is restricted to that range are constants, but then \(|\cos 0 - c| \leq 10^{-6}\) and \(|\cos \pi - c| \leq 10^{-6}\) \(2 = |1-(-1)| \leq |1-c| + |-1-c| \leq 2\cdot 10^{-6}\) contradiction.
4. True. \begin{align*} && (x^2-x^{-2})^2 &\geq 0 \\ \Leftrightarrow && x^4-2+x^{-4} &\geq0 \\ \Leftrightarrow && x^4+3+x^{-4} &\geq 5 \\ \end{align*}

Solution: We can always rotate the circle so that sides are parallel to the \(x\) and \(y\) axes. Therefore if one corner is \((a,b)\) the other coordinates are \((-a,b), (a,-b), (-a,-b)\) and the perimeter will be \(4(a+b)\). Therefore we wish to maximise \(4(a+b)\) subject to \(a^2+b^2 = \text{some constant}\). Notice that \(\frac{a+b}{2} \leq \sqrt{\frac{a^2+b^2}{2}}\) with equality when \(a = b\), therefore the maximum is a square. If \(n = 2\) then we are looking at \(2((2a)^2+(2b)^2) = 8(a^2+b^2)\) which is constant for all rectangles. If \(n=3\) we are maximising \(16(a^3+b^3) = 16(a^3+(c^2-a^2)^{3/2})\) which is maximised when \(a = 0, c\)

Solution: Suppose \(z = a+ib\), where \(a,b \in \mathbb{R}\) then the modulus of \(z\), \(|z| = \sqrt{a^2+b^2}\). Noting the similarity to the Pythagorean theorem, we can say that \(|z_1 - z_2|\) is the distance between \(z_1\) and \(z_2\) in the Argand diagram. \begin{align*} |z_1 + z_2| &= |(z_1 - 0) + (0 -z_2)| \\ &\underbrace{\leq}_{\text{the direct distance is shorter than going via }0} |z_1 - 0| + |0 - z_2| \\ &= |z_1| + |-z_2| \\ &= |z_1| + |z_2| \end{align*} Claim: \(\displaystyle \vert\,z_1+\cdots+z_n\,\vert \leq \sum_{i=1}^n |z_i|\) Proof: (By Induction) Base Case: \(n = 1, 2\) have been proven. Inductive step, suppose it is true for \(n = k\), then consider \(n = k+1\), ie \begin{align*} \vert\,z_1+\cdots+z_k+z_{k+1}\,\vert &\leq \vert\,z_1+\cdots+z_k\vert + \vert z_{k+1}\,\vert \\ &\underbrace{\leq}_{\text{inductive hypothesis}} \sum_{i=1}^k |z_i| + |z_{k+1}| \\ &= \sum_{i=1}^{k+1} |z_i| \end{align*} Therefore if our hypothesis is true for \(n = k\) it is true for \(n = k+1\), and so since it is true for \(n = 1\) it is true by the principle of mathematical induction for all integers \(n \geq 1\). Suppose \(|z| \leq 1/4\), then consider: \begin{align*} \vert a_1z+a_2z^2+\cdots +a_nz^n \vert &\leq \vert a_1 z\vert + \vert a_2z^2\vert + \cdots + \vert a_n z_n\ \vert \\ &= \vert a_1\vert\vert z\vert + \vert a_2\vert\vert z^2\vert + \cdots + \vert a_n\vert\vert z^n\ \vert \\ &\leq 3\left ( |z| + |z|^2 + \cdots + |z|^n \right) \\ &\leq 3 \left ( \frac{1}{4} + \frac1{4^2} + \cdots + \frac{1}{4^n} \right) \\ &< 3 \frac{1/4}{1-1/4} \\ &= 1 \end{align*} Therefore we cannot have equality and there are no solutions.

Solution: \begin{align*} && \E[\text{salary}] &= B\sum_{k=0}^N \frac{1}{1+k}\binom{N}{k}p^k(1-p)^{N-k} \\ \\ && (q+x)^N &= \sum_{k=0}^N \binom{N}{k}x^kq^{N-k} \\ \Rightarrow && \int_0^p(q+x)^N \d x &= \sum_{k=0}^N \binom{N}{k} \frac{p^{k+1}}{k+1}q^{N-k} \\ && \frac{(q+p)^{N+1}-q^{N+1}}{N+1} &= \frac{p}{B} \E[\text{salary}] \\ \Rightarrow && \E[\text{salary}] &= B\frac{1-q^{N+1}}{p(N+1)} \end{align*} Therefore if \(Ap(N+1) < B(1-(1-p)^{N+1})\) the expected value of the new salary is higher. (Whether or not the new salary is worth it in a risk adjusted sense is for the birds). We could model \(X\) by a Poisson random variable if \(N\) is large and \(Np = \lambda \) is small. Suppose \(X \approx Po(\lambda)\) then \begin{align*} \E \left [\frac{B}{1+X} \right] &= B\sum_{k=0}^\infty \frac{1}{1+k}\frac{e^{-\lambda}\lambda^k}{k!} \\ &= \frac{B}{\lambda} \sum_{k=0}^\infty e^{-\lambda} \frac{\lambda^{k+1}}{(k+1)!} \\ &= \frac{B}{\lambda}e^{-\lambda}(e^{\lambda}-1) \\ &= \frac{B(1-e^{-\lambda})}{\lambda} = B \frac{1-e^{-Np}}{Np} \end{align*}

Solution: \begin{align*} \text{COM}: && \mathbf{v}_1+\mathbf{v}_2 &= \mathbf{v}_1'+\mathbf{v}_2' \tag{1}\\ \text{COE}: && \mathbf{v}_1\cdot\mathbf{v}_1+\mathbf{v}_2\cdot\mathbf{v}_2 &= \mathbf{v}_1'\cdot\mathbf{v}_1'+\mathbf{v}_2'\cdot\mathbf{v}_2' \tag{2} \\ \\ (1): && (\mathbf{v}_1+\mathbf{v}_2 )\cdot(\mathbf{v}_1+\mathbf{v}_2 ) &= (\mathbf{v}_1'+\mathbf{v}_2' )\cdot(\mathbf{v}_1'+\mathbf{v}_2' ) \\ \Rightarrow && \mathbf{v}_1 \cdot \mathbf{v}_2 &= \mathbf{v}_1'\cdot \mathbf{v}_2' \\ && \text{Initial relative speed}^2 &= |\mathbf{v}_1 - \mathbf{v}_2|^2 \\ &&&= (\mathbf{v}_1 - \mathbf{v}_2) \cdot (\mathbf{v}_1 - \mathbf{v}_2) \\ &&&= \mathbf{v}_1\cdot \mathbf{v}_1 - 2 \mathbf{v}_1\cdot \mathbf{v}_2 + \mathbf{v}_2\cdot \mathbf{v}_2 \\ &&&= \mathbf{v}_1'\cdot\mathbf{v}_1'+\mathbf{v}_2'\cdot\mathbf{v}_2' -2 \mathbf{v}_1\cdot\mathbf{v}_2\\ &&&= \mathbf{v}_1'\cdot\mathbf{v}_1'+\mathbf{v}_2'\cdot\mathbf{v}_2' -2 \mathbf{v}_1'\cdot\mathbf{v}_2'\\ &&&= | \mathbf{v}_1'-\mathbf{v}_2'|^2 \\ &&&= \text{Final relative speed}^2 \end{align*} Since \(\mathbf{v}_1 \cdot 0 = 0\) we must have \(\mathbf{v}_1'\cdot\mathbf{v}_2' = \mathbf{v}_1\cdot0 = 0\) therefore their final velocities are perpendicular. We now must have \begin{align*} \text{COM}: && \mathbf{v}_1+\mathbf{v}_2 &= \mathbf{v}_1'+\mathbf{v}_2' \tag{3}\\ \Delta\text{E}: && (1-k)(\mathbf{v}_1\cdot\mathbf{v}_1+\mathbf{v}_2\cdot\mathbf{v}_2) &= \mathbf{v}_1'\cdot\mathbf{v}_1'+\mathbf{v}_2'\cdot\mathbf{v}_2' \tag{4} \\ \\ && 0 + \mathbf{v}_2 &= \mathbf{v}_1' + \mathbf{v}_2' \\ \Rightarrow && V^2 &= ( \mathbf{v}_1' + \mathbf{v}_2' ) \cdot ( \mathbf{v}_1' + \mathbf{v}_2' ) \\ &&&= \mathbf{v}_1'\cdot\mathbf{v}_1'+\mathbf{v}_2'\cdot\mathbf{v}_2' +2 \mathbf{v}_1'\cdot \mathbf{v}_2' \\ &&&= (1-k)V^2 + 2 (\mathbf{v}_2-\mathbf{v}_2') \cdot \mathbf{v}_2' \\ &&&= (1-k)V^2 + 2 \mathbf{v}_2 \cdot \mathbf{v}_2'-2\mathbf{v}_2'\cdot \mathbf{v}_2' \\ &&&= (1-k)V^2 + 2Vx \cos \theta - 2x^2 \\ \Rightarrow && 0 &= -kV^2 + 2Vx \cos \theta -2x^2 \\ \Delta \geq 0: && 0 &\leq 4V^2 \cos^2 \theta -8kV^2 \\ \Rightarrow && k &\leq \frac12\cos^2\theta \leq \frac12 \end{align*}

Solution: Plugging in \(x = i\) we obtain \((1-i)^4 = (-2i)^2 = -4 \Rightarrow b = -4\). \begin{align*} x^4(1-x)^4 &= x^4(1-4x+6x^2-4x^3+x^4) \\ &= x^8-4x^7+6x^6-4x^5+x^4 \\ &= x^6(x^2+1) - x^6 -4x^7+6x^6-4x^5+x^4 \\ &= x^6(x^2+1) -4x^5(x^2+1)+4x^5 +5x^6-4x^5+x^4 \\ &= (x^6-4x^5)(x^2+1) +5x^4(x^2+1)-5x^4+x^4 \\ &= (x^6-4x^5+5x^4)(x^2+1) -4x^2(x^2+1)+4x^2 \\ &= (x^6-4x^5+5x^4-4x^2)(x^2+1) +4(x^2+1)-4 \\ &= (x^6-4x^5+5x^4-4x^2+4)(x^2+1) -4 \\ \end{align*} So \begin{align*} \int_0^1 \frac{x^4(1-x)^4}{1+x^2} \d x &= \int_0^1 (x^6-4x^5+5x^4-4x^2+4) - \frac{4}{1+x^2} \d x \\ &= \frac17 - \frac46+1-\frac43+4 - \pi \\ &= \frac{22}7 - \pi \end{align*} \begin{align*} \int_0^1 x^4(1-x)^4 \d x &= B(5,5) \\ &= \frac{4!4!}{9!} \\ &= \frac1{630} \end{align*} Therefore since \(0 < \frac{x^4(1-x)^4}{1+x^2} < x^4(1-x)^4\) we must have that \begin{align*} && 0 &< \frac{22}7 - \pi \\ \Rightarrow && \pi & < \frac{22}{7} \\ && \frac{22}{7} - \pi &< \frac1{630} \\ \Rightarrow && \frac{22}{7} - \frac1{630} &< \pi \end{align*} which is what we wanted.

Solution: \begin{align*} && y &= \ln x - x \ln a \\ \Rightarrow && y' &= \frac1x - \ln a \\ && y'' &= -\frac{1}{x^2} \end{align*} Therefore the maximum is when \(x = \frac{1}{\ln a}\) and \(y_{max} = -\ln \ln a - 1\). If \(y_{max} < 0\) then \(y \neq 0\). But that's equivalent to \(a > e^{1/e}\). \begin{align*} && 0 &> -\ln \ln a - 1 \\ \Leftrightarrow && 1 &> - \ln \ln a \\ \Leftrightarrow && \ln \ln a &>-1 \\ \Leftrightarrow && \ln a &> e^{-1} \\ \Leftrightarrow && a & > e^{1/e} \end{align*} If \(0 < a < 1\) then, when \(x\) is small, \(\ln x - x \ln a\) is large and negative. When \(x\) is large and positive \(\ln x\) is positive and \(-x \ln a\) is positive. We also notice there is no turning point. Hence exactly one solution