Use the properties of logarithms, given that ln(2) ≈ 0.6931 and ln(3) ≈ 1.0986, to approximate the logarithm. Use a calculator to confirm your approximations. (Round your answers to four decimal places.) (a) ln(0.75) ≈ -0.5569 Incorrect: Your answer is incorrect. (b) ln(24) ≈ 8.3172 Incorrect: Your answer is incorrect. (c) ln( 3 18 ) ≈ 0.9635 Correct: Your answer is correct. (d) ln 1 72 ≈ 0.0458 Incorrect: Your answer is incorrect.

Answer:[tex]a) \ln(0.75) = -0.2876\\b) \ln(24) = 3.1779\\c) \ln(18)^\frac{1}{3}=0.9634\\d) \ln(\frac{1}{72}) = -4.2765[/tex]Step-by-step explanation:We are given that:[tex]ln(2) \approx 0.6931\\ln(3) \approx 1.0986[/tex]We have to approximate the logarithm.We use the following properties of log functions:[tex]\log(m\times n) = \log m + \log n\\\\\log(\displaystyle\frac{m}{n}) = \log m - \log n\\\\\log(m^n) = n\log m[/tex]1.  ln(0.75)[tex]\ln(0.75)\\\\=\ln(\displaystyle\frac{75}{100}) = \ln(\displaystyle\frac{3}{4}) = \ln 3 - \ln (2^2) = \ln 3 - 2\ln (2)\\\\= 1.0986 - 2(0.6931)\\= -0.2876[/tex]2. ln(24)[tex]\ln(24) \\=\ln(2\times 2\times 2\times 3) = \lm(2^3\times 3) = \ln(2^3) + \ln 3\\= 3\ln(2) + \ln(3) \\= 3(0.6931) + 1.0986\\= 3.1779[/tex]3. [tex]\ln(18)^\frac{1}{3}\\\\= \displaystyle\frac{1}{3}\ln (18)\\\\= \frac{1}{3}\ln(3^2\times 2)\\\\=\frac{1}{3}(2\ln 3 + \ln 2)\\\\=\frac{1}{3}(2(1.0986)+(0.6931))\\\\= 0.9634[/tex]4.[tex]\ln(\displaystyle\frac{1}{72})\\\\=\ln 1 - \ln 72\\= 0 - \ln(3^2\times 2^3)\\=-(2\ln(3) + 3\ln(2))\\=-(2( 1.0986)+3(0.6931))\\=-4.2765[/tex]The calculator approximations are:[tex]1. \ln(0.75) = -0.2876\\2. \ln(24) = 3.1780\\3. \ln(18)^\frac{1}{3}=0.9634\\4. \ln(\frac{1}{72}) = -4.2766[/tex]