Log Exp Resolucion E1575B

1. **Problema:** Resolver $\log_3 81 = x$ aplicando definición. La definición de logaritmo dice que $\log_a b = c$ es equivalente a $a^c = b$. $$\log_3 81 = x \implies 3^x = 81$$ Sabemos que $81 = 3^4$, entonces: $$3^x = 3^4$$ Por igualdad de potencias con la misma base: $$x = 4$$ 2. **Problema:** Resolver $\log_{\frac{1}{2}} 4 = x$. $$\log_{\frac{1}{2}} 4 = x \implies \left(\frac{1}{2}\right)^x = 4$$ Sabemos que $4 = 2^2$ y $\frac{1}{2} = 2^{-1}$, entonces: $$\left(2^{-1}\right)^x = 2^2$$ $$2^{-x} = 2^2$$ Igualando exponentes: $$-x = 2 \implies x = -2$$ 3. **Problema:** Resolver $\log_x \frac{1}{25} = -2$. $$\log_x \frac{1}{25} = -2 \implies x^{-2} = \frac{1}{25}$$ Sabemos que $\frac{1}{25} = 25^{-1} = 5^{-2}$, entonces: $$x^{-2} = 5^{-2}$$ Igualando bases: $$x = 5$$ 4. **Problema:** Resolver $\log_{\frac{1}{4}} \frac{1}{x} = -\frac{1}{2}$. $$\log_{\frac{1}{4}} \frac{1}{x} = -\frac{1}{2} \implies \left(\frac{1}{4}\right)^{-\frac{1}{2}} = \frac{1}{x}$$ $$\left(4^{-1}\right)^{-\frac{1}{2}} = \frac{1}{x}$$ $$4^{\frac{1}{2}} = \frac{1}{x}$$ $$2 = \frac{1}{x} \implies x = \frac{1}{2}$$ 5. **Problema:** Resolver $\log_{\frac{1}{16}} \frac{2x}{3} = -\frac{1}{2}$. $$\log_{\frac{1}{16}} \frac{2x}{3} = -\frac{1}{2} \implies \left(\frac{1}{16}\right)^{-\frac{1}{2}} = \frac{2x}{3}$$ $$\left(16^{-1}\right)^{-\frac{1}{2}} = \frac{2x}{3}$$ $$16^{\frac{1}{2}} = \frac{2x}{3}$$ $$4 = \frac{2x}{3}$$ Multiplicamos ambos lados por 3: $$3 \cdot 4 = 2x$$ $$12 = 2x$$ Dividimos entre 2: $$\cancel{2}x = \frac{12}{\cancel{2}}$$ $$x = 6$$ 6. **Problema:** Resolver $\log_4 x = 0$. $$\log_4 x = 0 \implies 4^0 = x$$ $$1 = x$$ 7. **Problema:** Resolver $\log_2 (x+1) = 0$. $$\log_2 (x+1) = 0 \implies 2^0 = x+1$$ $$1 = x+1$$ $$x = 0$$ 8. **Problema:** Resolver $\log_2 (x^2 + 3x - 2) = 1$. $$\log_2 (x^2 + 3x - 2) = 1 \implies 2^1 = x^2 + 3x - 2$$ $$2 = x^2 + 3x - 2$$ Pasamos todo a un lado: $$x^2 + 3x - 2 - 2 = 0$$ $$x^2 + 3x - 4 = 0$$ Factorizamos: $$(x + 4)(x - 1) = 0$$ Soluciones: $$x = -4, x = 1$$ Verificamos dominio: $x^2 + 3x - 2 > 0$ para que el logaritmo exista. Para $x=1$: $1 + 3 - 2 = 2 > 0$ válido. Para $x=-4$: $16 - 12 - 2 = 2 > 0$ válido. Ambas soluciones son válidas. 9. **Problema:** Resolver $\log_3 (8x + 9) = 4$. $$\log_3 (8x + 9) = 4 \implies 3^4 = 8x + 9$$ $$81 = 8x + 9$$ $$8x = 81 - 9 = 72$$ $$x = \frac{72}{8} = 9$$ 10. **Problema:** Resolver $\log_5 (13x - 18) = 3$. $$\log_5 (13x - 18) = 3 \implies 5^3 = 13x - 18$$ $$125 = 13x - 18$$ $$13x = 125 + 18 = 143$$ $$x = \frac{143}{13} = 11$$ --- 11. **Problema:** Calcular $\log_{45} 5 + 2 \log_{15} 3$ usando propiedades. Propiedades: - $a \log_b c = \log_b c^a$ - $\log_b m + \log_b n = \log_b (mn)$ si bases iguales. Primero, expresamos ambos logaritmos con base 15 o 45 para combinar. Pero bases diferentes, mejor evaluamos cada uno: $$2 \log_{15} 3 = \log_{15} 3^2 = \log_{15} 9$$ No se pueden sumar directamente, pero podemos usar cambio de base: $$\log_{45} 5 = \frac{\log 5}{\log 45}$$ $$\log_{15} 9 = \frac{\log 9}{\log 15}$$ No se simplifica fácilmente sin calculadora, dejamos así. 12. **Problema:** Calcular $2 \log 4 + \log 5 - \frac{1}{2} \log 6$. Usamos propiedades: $$2 \log 4 = \log 4^2 = \log 16$$ $$-\frac{1}{2} \log 6 = \log 6^{-\frac{1}{2}} = \log \frac{1}{\sqrt{6}}$$ Sumamos: $$\log 16 + \log 5 + \log \frac{1}{\sqrt{6}} = \log \left(16 \times 5 \times \frac{1}{\sqrt{6}}\right)$$ $$= \log \frac{80}{\sqrt{6}}$$ 13. **Problema:** Calcular $\log_{\frac{1}{2}} 20 - \frac{1}{3} \log_{\frac{1}{2}} 125$. $$\log_{\frac{1}{2}} 20 - \frac{1}{3} \log_{\frac{1}{2}} 125 = \log_{\frac{1}{2}} 20 - \log_{\frac{1}{2}} 125^{\frac{1}{3}}$$ $$= \log_{\frac{1}{2}} \frac{20}{5} = \log_{\frac{1}{2}} 4$$ Ya calculado antes, $\log_{\frac{1}{2}} 4 = -2$. 14. **Problema:** Calcular $\log_{\frac{1}{9}} 2 + \log_{\frac{1}{9}} 8 + 2 \log_{\frac{1}{9}} \frac{3}{4}$. $$2 \log_{\frac{1}{9}} \frac{3}{4} = \log_{\frac{1}{9}} \left(\frac{3}{4}\right)^2 = \log_{\frac{1}{9}} \frac{9}{16}$$ Sumamos: $$\log_{\frac{1}{9}} 2 + \log_{\frac{1}{9}} 8 + \log_{\frac{1}{9}} \frac{9}{16} = \log_{\frac{1}{9}} \left(2 \times 8 \times \frac{9}{16}\right)$$ $$= \log_{\frac{1}{9}} 9 = 1$$ --- 15. **Problema:** Resolver $3^x + 1 = 4$. $$3^x + 1 = 4 \implies 3^x = 3$$ $$3^x = 3^1$$ $$x = 1$$ 16. **Problema:** Resolver $3^{x+1} - 2 = 25$. $$3^{x+1} = 27$$ $$3^{x+1} = 3^3$$ Igualamos exponentes: $$x + 1 = 3 \implies x = 2$$ 17. **Problema:** Resolver $8^{x+1} = 2^{2x+7}$. $$8 = 2^3$$ $$8^{x+1} = (2^3)^{x+1} = 2^{3(x+1)} = 2^{3x + 3}$$ Igualamos exponentes: $$3x + 3 = 2x + 7$$ $$3x - 2x = 7 - 3$$ $$x = 4$$ 18. **Problema:** Resolver $5 \cdot 2^x + 2^x = 24$. Factorizamos $2^x$: $$(5 + 1) 2^x = 24$$ $$6 \cdot 2^x = 24$$ $$2^x = 4 = 2^2$$ $$x = 2$$ 19. **Problema:** Resolver $5^{x+3} - 5^{x+2} = 4$. Factorizamos $5^{x+2}$: $$5^{x+2} (5^1 - 1) = 4$$ $$5^{x+2} (5 - 1) = 4$$ $$5^{x+2} \cdot 4 = 4$$ $$5^{x+2} = 1 = 5^0$$ Igualamos exponentes: $$x + 2 = 0 \implies x = -2$$ 20. **Problema:** Resolver $8 \cdot 3^{x+1} + 3^{x+1} = 1$. Factorizamos $3^{x+1}$: $$(8 + 1) 3^{x+1} = 1$$ $$9 \cdot 3^{x+1} = 1$$ $$3^{x+1} = \frac{1}{9} = 3^{-2}$$ Igualamos exponentes: $$x + 1 = -2 \implies x = -3$$ 21. **Problema:** Resolver $2^{x+1} + 2^x = 12$. Factorizamos $2^x$: $$2^x (2 + 1) = 12$$ $$3 \cdot 2^x = 12$$ $$2^x = 4 = 2^2$$ $$x = 2$$