1. **Problem statement:** (a) Given the voltage-time relationship in a capacitive circuit: $$V = 210 \left(1 - e^{-\frac{R}{L}t}\right)$$ with $R=3\ \Omega$, $L=0.2\ \text{H}$, find the time $t$ when $V=110$ V. (b) Solve the equation: $$2\cosh x + \sinh x = 5$$ (c) Solve the equation: $$3\tan^3 x = \tan x$$ (d) Solve the logarithmic equation: $$\log_2(x) + \log_2(x+4) = 5$$ (e) Find the powers $m$ and $s$ such that acceleration $a$ depends on velocity $v$ and radius $r$ as: $$a \propto v^m r^s$$ --- 2. **Solution (a):** Given: $$V = 210 \left(1 - e^{-\frac{R}{L}t}\right),\quad V=110,\quad R=3,\quad L=0.2$$ Step 1: Substitute values: $$110 = 210 \left(1 - e^{-\frac{3}{0.2}t}\right)$$ Step 2: Divide both sides by 210: $$\frac{110}{210} = 1 - e^{-15t}$$ Step 3: Simplify fraction: $$\frac{11}{21} = 1 - e^{-15t}$$ Step 4: Rearrange to isolate exponential: $$e^{-15t} = 1 - \frac{11}{21} = \frac{10}{21}$$ Step 5: Take natural logarithm: $$-15t = \ln\left(\frac{10}{21}\right)$$ Step 6: Solve for $t$: $$t = -\frac{1}{15} \ln\left(\frac{10}{21}\right) = \frac{1}{15} \ln\left(\frac{21}{10}\right)$$ Step 7: Calculate numerical value: $$t \approx \frac{1}{15} \times 0.7419 = 0.04946\ \text{seconds}$$ --- 3. **Solution (b):** Equation: $$2\cosh x + \sinh x = 5$$ Recall: $$\cosh x = \frac{e^x + e^{-x}}{2}, \quad \sinh x = \frac{e^x - e^{-x}}{2}$$ Step 1: Substitute: $$2 \times \frac{e^x + e^{-x}}{2} + \frac{e^x - e^{-x}}{2} = 5$$ Step 2: Simplify: $$ (e^x + e^{-x}) + \frac{e^x - e^{-x}}{2} = 5$$ Step 3: Multiply both sides by 2: $$2(e^x + e^{-x}) + (e^x - e^{-x}) = 10$$ Step 4: Combine terms: $$2e^x + 2e^{-x} + e^x - e^{-x} = 10$$ Step 5: Simplify: $$3e^x + e^{-x} = 10$$ Step 6: Multiply both sides by $e^x$: $$3e^{2x} + 1 = 10 e^x$$ Step 7: Let $y = e^x$, then: $$3y^2 - 10 y + 1 = 0$$ Step 8: Solve quadratic: $$y = \frac{10 \pm \sqrt{100 - 12}}{6} = \frac{10 \pm \sqrt{88}}{6} = \frac{10 \pm 2\sqrt{22}}{6}$$ Step 9: Calculate roots: $$y_1 = \frac{10 + 2\sqrt{22}}{6}, \quad y_2 = \frac{10 - 2\sqrt{22}}{6}$$ Step 10: Since $y = e^x > 0$, both positive roots are valid. Step 11: Find $x$: $$x = \ln y$$ So, $$x_1 = \ln\left(\frac{10 + 2\sqrt{22}}{6}\right), \quad x_2 = \ln\left(\frac{10 - 2\sqrt{22}}{6}\right)$$ --- 4. **Solution (c):** Equation: $$3 \tan^3 x = \tan x$$ Step 1: Rearrange: $$3 \tan^3 x - \tan x = 0$$ Step 2: Factor out $\tan x$: $$\tan x (3 \tan^2 x - 1) = 0$$ Step 3: Set each factor to zero: (i) $\tan x = 0 \implies x = n\pi, n \in \mathbb{Z}$ (ii) $3 \tan^2 x - 1 = 0 \implies \tan^2 x = \frac{1}{3} \implies \tan x = \pm \frac{1}{\sqrt{3}}$ Step 4: Solve for $x$: $$x = \arctan\left(\pm \frac{1}{\sqrt{3}}\right) + n\pi = \pm \frac{\pi}{6} + n\pi, n \in \mathbb{Z}$$ --- 5. **Solution (d):** Equation: $$\log_2(x) + \log_2(x+4) = 5$$ Step 1: Use log property: $$\log_2(x(x+4)) = 5$$ Step 2: Rewrite as exponential: $$x(x+4) = 2^5 = 32$$ Step 3: Expand: $$x^2 + 4x = 32$$ Step 4: Rearrange: $$x^2 + 4x - 32 = 0$$ Step 5: Solve quadratic: $$x = \frac{-4 \pm \sqrt{16 + 128}}{2} = \frac{-4 \pm \sqrt{144}}{2} = \frac{-4 \pm 12}{2}$$ Step 6: Roots: $$x_1 = \frac{8}{2} = 4, \quad x_2 = \frac{-16}{2} = -8$$ Step 7: Domain check: $x > 0$ and $x+4 > 0$ implies $x > 0$ and $x > -4$, so $x=4$ valid, $x=-8$ invalid. Final solution: $$x=4$$ --- 6. **Solution (e):** Given: $$a \propto v^m r^s$$ Step 1: Write dimensions: $$[a] = L T^{-2}, \quad [v] = L T^{-1}, \quad [r] = L$$ Step 2: Substitute: $$L T^{-2} = (L T^{-1})^m \times L^s = L^{m+s} T^{-m}$$ Step 3: Equate powers of $L$ and $T$: For $L$: $$1 = m + s$$ For $T$: $$-2 = -m \implies m = 2$$ Step 4: Substitute $m=2$ into $L$ equation: $$1 = 2 + s \implies s = -1$$ Final result: $$a \propto v^2 r^{-1}$$ --- **Summary:** (a) $$t = \frac{1}{15} \ln\left(\frac{21}{10}\right) \approx 0.0495\ \text{s}$$ (b) $$x = \ln\left(\frac{10 \pm 2\sqrt{22}}{6}\right)$$ (c) $$x = n\pi, \quad x = \pm \frac{\pi}{6} + n\pi, \quad n \in \mathbb{Z}$$ (d) $$x=4$$ (e) $$a \propto v^2 r^{-1}$$