Question

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 1

Answer:

$a) \ln(0.75) = -0.2876\\b) \ln(24) = 3.1779\\c) \ln(18)^\frac{1}{3}=0.9634\\d) \ln(\frac{1}{72}) = -4.2765$

Step-by-step explanation:

We are given that:

$ln(2) \approx 0.6931\\ln(3) \approx 1.0986$

We have to approximate the logarithm.

We use the following properties of log functions:

$\log(m\times n) = \log m + \log n\\\\\log(\displaystyle\frac{m}{n}) = \log m - \log n\\\\\log(m^n) = n\log m$

1.  ln(0.75)

$\ln(0.75)\\\\=\ln(\displaystyle\frac{75}{100}) = \ln(\displaystyle\frac{3}{4}) = \ln 3 - \ln (2^2) = \ln 3 - 2\ln (2)\\\\= 1.0986 - 2(0.6931)\\= -0.2876$

2. ln(24)

$\ln(24) \\=\ln(2\times 2\times 2\times 3) = \lm(2^3\times 3) = \ln(2^3) + \ln 3\\= 3\ln(2) + \ln(3) \\= 3(0.6931) + 1.0986\\= 3.1779$

3.

$\ln(18)^\frac{1}{3}\\\\= \displaystyle\frac{1}{3}\ln (18)\\\\= \frac{1}{3}\ln(3^2\times 2)\\\\=\frac{1}{3}(2\ln 3 + \ln 2)\\\\=\frac{1}{3}(2(1.0986)+(0.6931))\\\\= 0.9634$

4.

$\ln(\displaystyle\frac{1}{72})\\\\=\ln 1 - \ln 72\\= 0 - \ln(3^2\times 2^3)\\=-(2\ln(3) + 3\ln(2))\\=-(2( 1.0986)+3(0.6931))\\=-4.2765$

The calculator approximations are:

$1. \ln(0.75) = -0.2876\\2. \ln(24) = 3.1780\\3. \ln(18)^\frac{1}{3}=0.9634\\4. \ln(\frac{1}{72}) = -4.2766$

Answer 2

Answer:

Since, logarithm properties are,

$\ln a.b = \ln a + \ln b$

$\ln a^b = b\ln a$

$\ln (\frac{a}{b}) = \ln a - \ln b$

Given,

ln(2) ≈ 0.6931 and ln(3) ≈ 1.0986,

(a) ln(0.75) = ln(3/4)

= ln(3) - ln(4)

= ln(3) - ln(2²)

= ln(3) - 2ln (2)

= 1.0986 - 2(0.6931)

= -0.2876,

(b) ln(24) = ln(6 × 4)

= ln (6) + ln (4)

= ln (3 × 2) + ln(2²)

= ln (3) + ln(2) + 2 ln(2)

= ln 3 + 3 ln 2

= 1.0986 + 3(0.6931)

= 3.1779,

(c) ln (∛ 18)

$= \ln (18)^\frac{1}{3}$

$=\frac{1}{3}\ln (18)$

$=\frac{1}{3}(\ln (3\times 2\times 3))$

$=\frac{1}{3}(\ln 3 + \ln 2 + \ln 3)$

$=\frac{1}{3}(1.0986+0.6931+1.0986)$

$=\frac{1.0986+0.6931+1.0986}{3}$

≈ 0.9634

(d) $\ln (\frac{1}{72})$

$=\ln 1 - \ln 72$

$=0 - \ln ( 36 × 2)$

$=-\ln 36 - \ln 2$

$=-\ln 6^2 - \ln 2$

$= -2\ln 6 - \ln 2$

$=-2(\ln (3\times 2)) - \ln 2$

$=-2\ln 3 - 2\ln 2 - \ln 2$

$=-2\ln 3 - 3\ln 2$

$=-2(1.0986) - 3(0.6931)$

= -4.2765