1. El problema nos pide expresar varios logaritmos en términos de $a=\log 2$, $b=\log 3$ y $c=\log 5$. 2. Recordemos las propiedades importantes de los logaritmos: - $\log(xy) = \log x + \log y$ - $\log\left(\frac{x}{y}\right) = \log x - \log y$ - $\log(x^n) = n \log x$ - Cambio de base: $\log_m n = \frac{\log n}{\log m}$ 3. Ahora, resolvemos cada uno: (i) $\log 30 = \log(2 \times 3 \times 5) = \log 2 + \log 3 + \log 5 = a + b + c$ (ii) $\log \left(\frac{15}{2}\right) = \log 15 - \log 2 = \log(3 \times 5) - \log 2 = (b + c) - a = b + c - a$ (iii) $\log 125 = \log(5^3) = 3 \log 5 = 3c$ (iv) $\log \left(\frac{27}{8}\right) = \log 27 - \log 8 = \log(3^3) - \log(2^3) = 3b - 3a = 3(b - a)$ (v) $\log 0.000002 = \log \left(2 \times 10^{-6}\right) = \log 2 + \log 10^{-6} = a + (-6) \log 10 = a - 6$ (vi) $\log_5 15 = \frac{\log 15}{\log 5} = \frac{\log(3 \times 5)}{c} = \frac{b + c}{c} = \frac{b}{c} + 1$ (vii) $\log_3 2 = \frac{\log 2}{\log 3} = \frac{a}{b}$ Respuesta final: (i) $a + b + c$ (ii) $b + c - a$ (iii) $3c$ (iv) $3(b - a)$ (v) $a - 6$ (vi) $\frac{b}{c} + 1$ (vii) $\frac{a}{b}$