1. Problem: Evaluate the numerical value of the logarithmic expressions. (a) Calculate $\log_2 \sqrt{8} + \log_3 \frac{1}{3} + 6^{\log_6 0.1}$. - Use $\log_b a^c = c \log_b a$ and $\sqrt{8} = 8^{1/2}$. - $\log_2 \sqrt{8} = \log_2 8^{1/2} = \frac{1}{2} \log_2 8 = \frac{1}{2} \times 3 = 1.5$. - $\log_3 \frac{1}{3} = \log_3 3^{-1} = -1$. - $6^{\log_6 0.1} = 0.1$ because $a^{\log_a b} = b$. - Sum: $1.5 + (-1) + 0.1 = 0.6$. (b) Calculate $\log_{1/3} 27 + 2 \log_2 4^2 + \log_5 \sqrt{5} - 2 \log_5 1$. - $\log_{1/3} 27 = \frac{\log 27}{\log (1/3)}$; since $27=3^3$ and $1/3=3^{-1}$, this equals $\frac{3 \log 3}{-1 \log 3} = -3$. - $2 \log_2 4^2 = 2 \log_2 16 = 2 \times 4 = 8$. - $\log_5 \sqrt{5} = \log_5 5^{1/2} = \frac{1}{2}$. - $\log_5 1 = 0$, so $-2 \times 0 = 0$. - Sum: $-3 + 8 + 0.5 + 0 = 5.5$. 2. Problem: Solve logarithmic equations. (a) $\log (2x - 5) + \log (x - 1) = 2 \log 2 + \log 5$. - Use $\log a + \log b = \log (ab)$ and $k \log a = \log a^k$. - Left: $\log ((2x - 5)(x - 1))$. - Right: $\log 2^2 + \log 5 = \log (4) + \log 5 = \log (20)$. - So, $(2x - 5)(x - 1) = 20$. - Expand: $2x^2 - 2x - 5x + 5 = 20 \Rightarrow 2x^2 - 7x + 5 = 20$. - Simplify: $2x^2 - 7x - 15 = 0$. - Solve quadratic: $x = \frac{7 \pm \sqrt{49 + 120}}{4} = \frac{7 \pm 13}{4}$. - Solutions: $x=5$ or $x=-1.5$. - Check domain: $2x-5>0$ and $x-1>0$ implies $x>2.5$ and $x>1$, so $x=5$ valid, $x=-1.5$ invalid. (b) $\log_2 (x + 3) + \log_2 (2x + 4) - \log_{11} 121 = 0$. - $\log_2 (x + 3) + \log_2 (2x + 4) = \log_2 ((x+3)(2x+4))$. - $\log_{11} 121 = \log_{11} 11^2 = 2$. - Equation: $\log_2 ((x+3)(2x+4)) - 2 = 0 \Rightarrow \log_2 ((x+3)(2x+4)) = 2$. - So, $(x+3)(2x+4) = 2^2 = 4$. - Expand: $2x^2 + 4x + 6x + 12 = 4 \Rightarrow 2x^2 + 10x + 12 = 4$. - Simplify: $2x^2 + 10x + 8 = 0$. - Divide by 2: $x^2 + 5x + 4 = 0$. - Factor: $(x+4)(x+1) = 0$. - Solutions: $x = -4$ or $x = -1$. - Check domain: $x+3>0 \Rightarrow x>-3$, $2x+4>0 \Rightarrow x>-2$. - Both solutions invalid since $-4 < -3$ and $-1 > -3$ but $2(-1)+4=2>0$ so $x=-1$ valid. (c) $3 \log_2 2 + \log_a (x - 2) = \log_a 12(x + 3) - \log_a 3$. - $3 \log_2 2 = 3 \times 1 = 3$. - Use $\log_a m - \log_a n = \log_a \frac{m}{n}$. - Equation: $3 + \log_a (x-2) = \log_a \frac{12(x+3)}{3} = \log_a 4(x+3)$. - Rearrange: $\log_a (x-2) - \log_a 4(x+3) = -3$. - $\log_a \frac{x-2}{4(x+3)} = -3$. - Rewrite: $\frac{x-2}{4(x+3)} = a^{-3} = \frac{1}{a^3}$. - Cross multiply: $a^3 (x-2) = 4(x+3)$. - Solve for $x$: $a^3 x - 2 a^3 = 4x + 12$. - $x(a^3 - 4) = 2 a^3 + 12$. - $x = \frac{2 a^3 + 12}{a^3 - 4}$. (d) $\log_3 (x + 4) = \log_7 7 - \log_3 (x + 2)$. - $\log_7 7 = 1$. - Rearrange: $\log_3 (x + 4) + \log_3 (x + 2) = 1$. - $\log_3 ((x+4)(x+2)) = 1$. - So, $(x+4)(x+2) = 3^1 = 3$. - Expand: $x^2 + 6x + 8 = 3$. - Simplify: $x^2 + 6x + 5 = 0$. - Factor: $(x+5)(x+1) = 0$. - Solutions: $x = -5$ or $x = -1$. - Check domain: $x+4>0 \Rightarrow x>-4$, $x+2>0 \Rightarrow x>-2$. - $x=-5$ invalid, $x=-1$ valid. (e) $\log (x + 3) + \log (2x - 7) = \log (2x - 1)$. - Combine left: $\log ((x+3)(2x-7)) = \log (2x - 1)$. - So, $(x+3)(2x-7) = 2x - 1$. - Expand: $2x^2 - 7x + 6x - 21 = 2x - 1$. - Simplify: $2x^2 - x - 21 = 2x - 1$. - Rearrange: $2x^2 - x - 21 - 2x + 1 = 0 \Rightarrow 2x^2 - 3x - 20 = 0$. - Solve quadratic: $x = \frac{3 \pm \sqrt{9 + 160}}{4} = \frac{3 \pm 13}{4}$. - Solutions: $x=4$ or $x=-2.5$. - Check domain: $x+3>0 \Rightarrow x>-3$, $2x-7>0 \Rightarrow x>3.5$, $2x-1>0 \Rightarrow x>0.5$. - Valid solution: $x=4$. (f) $2^x = 3^{2x - 1}$. - Take $\log$ both sides: $x \log 2 = (2x - 1) \log 3$. - Expand: $x \log 2 = 2x \log 3 - \log 3$. - Rearrange: $x \log 2 - 2x \log 3 = - \log 3$. - Factor: $x (\log 2 - 2 \log 3) = - \log 3$. - $x = \frac{- \log 3}{\log 2 - 2 \log 3}$. 3. Problem: Given $\log_n u = 7$ and $\log_n w = -9$, find $\log_n \left( \frac{n^2 w^4}{u^5} \right)$. - Use $\log_n (a/b) = \log_n a - \log_n b$ and $\log_n (a b) = \log_n a + \log_n b$. - $\log_n \left( n^2 w^4 / u^5 \right) = \log_n n^2 + \log_n w^4 - \log_n u^5$. - $= 2 + 4 \log_n w - 5 \log_n u = 2 + 4(-9) - 5(7) = 2 - 36 - 35 = -69$. 4. Problem: Express $\log_2 \left( \frac{\sqrt{x}(x-5)}{x^3} \right)$ as sum/difference/product of logarithms. - Rewrite: $\log_2 \left( \frac{x^{1/2} (x-5)}{x^3} \right) = \log_2 \left( x^{1/2} (x-5) x^{-3} \right)$. - Simplify: $\log_2 \left( x^{1/2 - 3} (x-5) \right) = \log_2 \left( x^{-5/2} (x-5) \right)$. - Use $\log (ab) = \log a + \log b$: $\log_2 (x-5) + \log_2 x^{-5/2}$. - Use $\log a^c = c \log a$: $\log_2 (x-5) - \frac{5}{2} \log_2 x$. 5. Problem: Given $2 \log_a 8 + \log_a (1/a)^5 = 19$, find base $a$. - $2 \log_a 8 = \log_a 8^2 = \log_a 64$. - $\log_a (1/a)^5 = \log_a a^{-5} = -5$. - Equation: $\log_a 64 - 5 = 19 \Rightarrow \log_a 64 = 24$. - Rewrite: $a^{24} = 64 = 2^6$. - So, $a^{24} = 2^6$. - Take $24$th root: $a = 2^{6/24} = 2^{1/4}$. 6. Problem: Identify which function could represent increasing logarithmic function $f$ with domain $]5, +\infty[$ passing through $(9,10)$. - Check domain and increasing behavior. - (A) $f(x) = 3 \log_2 (x - 5) + 4$; domain $x>5$, base 2 >1 increasing. - Evaluate at $x=9$: $3 \log_2 4 + 4 = 3 \times 2 + 4 = 10$ correct. - (B) $5 \log_{1/2} (x - 1) + 5$; base $1/2 < 1$ decreasing, so no. - (C) $\log_4 (x - 5) + 8$; domain $x>5$, base 4 >1 increasing. - Evaluate at $x=9$: $\log_4 4 + 8 = 1 + 8 = 9 \neq 10$. - (D) $15 \log_{14} (x + 5) - 5$; domain $x > -5$, not $>5$. - So, function (A) fits. 7. Problem: Solve $5^{8x - 12} = 25^{2x + 12}$. - Rewrite $25 = 5^2$. - Equation: $5^{8x - 12} = (5^2)^{2x + 12} = 5^{4x + 24}$. - Equate exponents: $8x - 12 = 4x + 24$. - Solve: $8x - 4x = 24 + 12 \Rightarrow 4x = 36 \Rightarrow x = 9$. 8. Problem: Find inverse $f^{-1}(x)$ of $f(x) = 2 \times 5^{x + 1}$. - Write $y = 2 \times 5^{x + 1}$. - Divide: $\frac{y}{2} = 5^{x + 1}$. - Take $\log_5$: $\log_5 \frac{y}{2} = x + 1$. - Solve for $x$: $x = \log_5 \frac{y}{2} - 1$. - So, $f^{-1}(x) = \log_5 \frac{x}{2} - 1$. Final answers: (a) 0.6 (b) 5.5 (c) $x=5$ (d) $x=-1$ (e) $x=4$ (f) $x=\frac{-\log 3}{\log 2 - 2 \log 3}$ (g) $-69$ (h) $\log_2 (x-5) - \frac{5}{2} \log_2 x$ (i) $a = 2^{1/4}$ (j) Function (A) (k) $x=9$ (l) $f^{-1}(x) = \log_5 \frac{x}{2} - 1$