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

Since, logarithm properties are,

`\ln a.b = \ln a + \ln b`

`\ln a^b = b\ln a`

`\ln ((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)^(1)/(3)`

`=(1)/(3)\ln (18)`

`=(1)/(3)(\ln (3* 2* 3))`

`=(1)/(3)(\ln 3 + \ln 2 + \ln 3)`

`=(1)/(3)(1.0986+0.6931+1.0986)`

`=(1.0986+0.6931+1.0986)/(3)`

≈ 0.9634

(d)
`\ln ((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* 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

Answer 2

`a) \ln(0.75) = -0.2876\\b) \ln(24) = 3.1779\\c) \ln(18)^(1)/(3)=0.9634\\d) \ln((1)/(72)) = -4.2765`

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* n) = \log m + \log n\\\\\log(\displaystyle(m)/(n)) = \log m - \log n\\\\\log(m^n) = n\log m`

1. ln(0.75)

`\ln(0.75)\\\\=\ln(\displaystyle(75)/(100)) = \ln(\displaystyle(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* 2* 2* 3) = \lm(2^3* 3) = \ln(2^3) + \ln 3\\= 3\ln(2) + \ln(3) \\= 3(0.6931) + 1.0986\\= 3.1779`

3.

`\ln(18)^(1)/(3)\\\\= \displaystyle(1)/(3)\ln (18)\\\\= (1)/(3)\ln(3^2* 2)\\\\=(1)/(3)(2\ln 3 + \ln 2)\\\\=(1)/(3)(2(1.0986)+(0.6931))\\\\= 0.9634`

4.

`\ln(\displaystyle(1)/(72))\\\\=\ln 1 - \ln 72\\= 0 - \ln(3^2* 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)^(1)/(3)=0.9634\\4. \ln((1)/(72)) = -4.2766`