1. Given $\log_{10} 2 = m$ and $\log_{10} 3 = n$, evaluate $\log_{10} 24$. We use the property that $\log(ab) = \log a + \log b$. Express 24 as $24 = 2^3 \times 3$. So, $\log_{10} 24 = \log_{10} (2^3 \times 3) = \log_{10} 2^3 + \log_{10} 3 = 3 \log_{10} 2 + \log_{10} 3 = 3m + n$. 2. Given $\log_{10} 3 = a$ and $\log_{10} 5 = b$, evaluate $\log_{10} 75$. Express 75 as $75 = 3 \times 25 = 3 \times 5^2$. So, $\log_{10} 75 = \log_{10} 3 + \log_{10} 5^2 = a + 2b$. 3. Evaluate $\log y = 3\log 2 + \log 3 - \log 6$. Use log rules: $\log a^b = b \log a$, and $\log (a) - \log (b) = \log (\frac{a}{b})$. Rewrite as $\log y = \log 2^3 + \log 3 - \log 6 = \log 8 + \log 3 - \log 6 = \log (\frac{8 \times 3}{6}) = \log 4$. So, $y = 4$. 4. Evaluate $\frac{\log_3 9 - \log_2 8}{\log_3 81}$. Calculate each log: $\log_3 9 = 2$ (since $3^2=9$), $\log_2 8=3$ (since $2^3=8$), $\log_3 81=4$ (since $3^4=81$). So expression is $\frac{2 - 3}{4} = \frac{-1}{4} = -\frac{1}{4}$. 5. Solve $\log_{10}(8x+7) - \log_{10}(2x+1) = \log_{10}(x+2)$. Use property $\log a - \log b = \log (\frac{a}{b})$. Rewrite as $\log_{10} \left( \frac{8x+7}{2x+1} \right) = \log_{10} (x+2)$. Equate arguments: $\frac{8x+7}{2x+1} = x+2$. Multiply both sides by $2x+1$: $8x+7 = (x+2)(2x+1) = 2x^2 + 5x + 2$. Rearranging: $0 = 2x^2 + 5x + 2 - 8x - 7 = 2x^2 - 3x - 5$. Solve quadratic $2x^2 - 3x - 5 = 0$. Using quadratic formula $x = \frac{3 \pm \sqrt{9 + 40}}{4} = \frac{3 \pm \sqrt{49}}{4}$. So $x = \frac{3 \pm 7}{4}$. Two solutions: $x=\frac{10}{4}=2.5$ and $x=\frac{-4}{4}=-1$. Check domain: must have $8x+7 >0$, $2x+1 >0$, and $x+2 >0$. At $x=2.5$, all positive. At $x=-1$, $8(-1)+7 = -1<0$ (invalid). So solution: $x=2.5$. 6. Evaluate $\log_{10} (\frac{75}{10}) - 2 \log_{10} (\frac{5}{9}) + \log_{10} (\frac{100}{243})$. First simplify each: $\log_{10} 7.5 - 2 \log_{10} \frac{5}{9} + \log_{10} \frac{100}{243}$. Rewrite all as: $\log_{10} 7.5 + \log_{10} (\frac{5}{9})^{-2} + \log_{10} \frac{100}{243}$. Use exponent inside log inverse: $(\frac{5}{9})^{-2} = (\frac{9}{5})^{2} = \frac{81}{25}$. Expression becomes: $\log_{10} 7.5 + \log_{10} \frac{81}{25} + \log_{10} \frac{100}{243}$. Combine logs: $\log_{10} \left( 7.5 \times \frac{81}{25} \times \frac{100}{243} \right)$. Calculate inside the log: $7.5 = \frac{15}{2}$, $\frac{15}{2} \times \frac{81}{25} = \frac{15 \times 81}{2 \times 25} = \frac{1215}{50}$. Multiply by $\frac{100}{243}$: $\frac{1215}{50} \times \frac{100}{243} = \frac{1215 \times 100}{50 \times 243}$. Simplify: $\frac{121500}{12150} = 10$. So log expression is $\log_{10} 10 = 1$. 7. Solve $3 \log x + \log 3 = \log 81$, where base is 10. Use $3 \log x = \log x^3$ and $\log 81 = \log 3^4$. Rewrite: $\log x^3 + \log 3 = \log 3^4$. Combine: $\log (3x^3) = \log 81$. Equate arguments: $3x^3 = 81$. Divide both sides: $x^3 = 27$. So $x = 3$. 8. Solve $\log (5x-4) = \log (x+1) + \log 4$. Use $\log a + \log b = \log(ab)$. Rewrite: $\log (5x-4) = \log (4(x+1))$. Equate arguments: $5x - 4 = 4x + 4$. Simplify: $x = 8$. Check domain: $5(8)-4=36>0$, $8+1=9>0$ valid. 9. Solve $16 x^{2x+1} = 4 \times 8^{1-x}$. Rewrite $16 = 2^4$, $4=2^2$, $8=2^3$. Expression becomes: $2^4 x^{2x+1} = 2^2 (2^3)^{1-x}$. Simplify right side exponent: $2^2 \times 2^{3(1-x)} = 2^{2 + 3 - 3x} = 2^{5 - 3x}$. So $2^4 x^{2x+1} = 2^{5-3x}$. Divide both sides by $2^4$: $x^{2x+1} = 2^{1 - 3x}$. Rewrite RHS as $x^{2x+1} = (2^{\frac{1 - 3x}{\log_2 x}})$ but it is nontrivial to solve algebraically, so let's try substitution: assume $x=2^k$. Then $x^{2x+1} = (2^k)^{2(2^k)+1} = 2^{k(2(2^k)+1)}$, RHS is $2^{1 - 3(2^k)}$. Equate exponents: $k(2(2^k)+1) = 1 - 3(2^k)$. Try $k=0$: LHS = 0, RHS = 1 - 3(1)= -2 no. Try $k=-1$: $x=2^{-1} = 0.5$. LHS: $-1(2(0.5)+1) = -1(1+1) = -2$. RHS: $1 - 3(0.5) = 1 - 1.5 = -0.5$. No. Try $k=1$: $x=2$. LHS: $1(2(2)+1) = 1(4+1) = 5$. RHS: $1 - 3(2)=1-6=-5$. No. Try $k=-0.5$: approximate $x=2^{-0.5}=1/\sqrt{2}\approx0.707$, LHS: $-0.5(2(0.707)+1) = -0.5(2.414)=-1.207$, RHS: $1-3(0.707)=1-2.121=-1.121$ close but no. No easy closed form; leave as implicit solution or numerically approx. 10. Solve $2^{\sqrt{2x+1}} = 32$. Rewrite $32=2^5$. Equate exponents: $\sqrt{2x+1} = 5$. Square both sides: $2x+1=25$. Solve: $2x=24$, $x=12$. 11. Given $3^m = 81$, solve for $m$. Rewrite $81 = 3^4$. So $3^m=3^4$ gives $m=4$. 12. Solve $2^x + 2^{x-1} = 48$. Rewrite: $2^{x-1} = \frac{2^x}{2}$. So equation becomes $2^x + \frac{2^x}{2} = 48$. Combine terms: $2^x \left(1 + \frac{1}{2} \right) = 48$. So $2^x \times \frac{3}{2} = 48$. Solve: $2^x = \frac{48 \times 2}{3} = 32$. Rewrite $32 = 2^5$. Hence $x=5$. 13. Solve $3 \log (2+x) = \log 27$. Rewrite: $3 \log (2+x) = \log (2+x)^3$. Rewrite $27 = 3^3$. So $\log (2+x)^3 = \log 3^3$. Equate arguments: $(2+x)^3 = 3^3$. Take cube root: $2+x = 3$, so $x=1$.