University:
Morehouse College
Course:
HMTH 105 | College Algebra 1
Academic year:

2024
Views:

403

Pages:

4
Author:

Reynaldo Wong

Question 1 Solve for 𝑥 x using log ⁡ 50 = 600 𝑒 − 0.4 𝑥 log50=600e −0.4x . Answer 1 Step 1: Solve the equation: 50 = 600 𝑒 − 0.4 𝑥 50=600e −0.4x Divide both sides by 600: 50 600 = 𝑒 − 0.4 𝑥 600 50 =e −0.4x 0.0833 = 𝑒 − 0.4 𝑥 0.0833=e −0.4x Take the natural logarithm (ln) of both sides: ln ⁡ ( 0.0833 ) = − 0.4 𝑥 ln(0.0833)=−0.4x 𝑥 = − ln ⁡ ( 0.0833 ) 0.4 x= 0.4 −ln(0.0833) Step 2: Calculate: ln ⁡ ( 0.0833 ) ≈ − 2.485 ln(0.0833)≈−2.485 𝑥 = − ( − 2.485 ) 0.4 = 2.485 0.4 = 6.2125 x= 0.4 −(−2.485) = 0.4 2.485 =6.2125 Result: 𝑥 ≈ 6.32 x≈6.32 Question 2 Use the properties of logarithms, given that ln ⁡ ( 2 ) = − 0.6931 ln(2)=−0.6931 and ln ⁡ ( 3 ) = 1.0986 ln(3)=1.0986, to approximate the logarithm. Use a calculator to confirm your approximations. Answer 1 a) ln ⁡ ( 0.75 ) ln(0.75): Using ln ⁡ ( 𝑎 / 𝑏 ) = ln ⁡ ( 𝑎 ) − ln ⁡ ( 𝑏 ) ln(a/b)=ln(a)−ln(b): ln ⁡ ( 0.75 ) = ln ⁡ ( 3 4 ) = ln ⁡ ( 3 ) − ln ⁡ ( 4 ) ln(0.75)=ln( 4 3 )=ln(3)−ln(4) ln ⁡ ( 4 ) = ln ⁡ ( 2 2 ) = 2 ln ⁡ ( 2 ) = 2 ( − 0.6931 ) = − 1.3862 ln(4)=ln(2 2 )=2ln(2)=2(−0.6931)=−1.3862 ln ⁡ ( 0.75 ) = ln ⁡ ( 3 ) − ln ⁡ ( 4 ) = 1.0986 − ( − 1.3862 ) = − 0.2876 ln(0.75)=ln(3)−ln(4)=1.0986−(−1.3862)=−0.2876 b) ln ⁡ ( 54 ) ln(54): Using ln ⁡ ( 𝑎 𝑏 ) = ln ⁡ ( 𝑎 ) + ln ⁡ ( 𝑏 ) ln(ab)=ln(a)+ln(b): ln ⁡ ( 54 ) = ln ⁡ ( 9 ⋅ 6 ) = ln ⁡ ( 9 ) + ln ⁡ ( 6 ) ln(54)=ln(9⋅6)=ln(9)+ln(6) ln ⁡ ( 9 ) = ln ⁡ ( 3 2 ) = 2 ln ⁡ ( 3 ) = 2 ( 1.0986 ) = 2.1972 ln(9)=ln(3 2 )=2ln(3)=2(1.0986)=2.1972 ln ⁡ ( 6 ) = ln ⁡ ( 2 ⋅ 3 ) = ln ⁡ ( 2 ) + ln ⁡ ( 3 ) = − 0.6931 + 1.0986 = 0.4055 ln(6)=ln(2⋅3)=ln(2)+ln(3)=−0.6931+1.0986=0.4055 ln ⁡ ( 54 ) = 2.1972 + 0.4055 = 3.9889 ln(54)=2.1972+0.4055=3.9889 c) ln ⁡ ( 312 ) ln( 312 ): ln ⁡ ( 312 ) = ln ⁡ ( 31 2 1 / 2 ) = 1 2 ln ⁡ ( 312 ) ln( 312 )=ln(312 1/2 )= 2 1 ln(312) Break 312 = 3 ⋅ 4 ⋅ 26 312=3⋅4⋅26, then: ln ⁡ ( 312 ) = ln ⁡ ( 3 ) + ln ⁡ ( 4 ) + ln ⁡ ( 26 ) ln(312)=ln(3)+ln(4)+ln(26) Simplify further to approximate. Question 3 Calculate the ratio of NaF to HF required to create a buffer with pH = 4.00 pH=4.00. Answer 1 From the Henderson-Hasselbalch equation: pH = pKa + log ⁡ ( [A − ] [HA] ) pH=pKa+log( [HA] [A − ] ) 4.00 = 3.60 + log ⁡ ( [NaF] [HF] ) 4.00=3.60+log( [HF] [NaF] ) log ⁡ ( [NaF] [HF] ) = 4.00 − 3.60 = 0.40 log( [HF] [NaF] )=4.00−3.60=0.40 Convert to ratio: [NaF] [HF] = 1 0 0.40 ≈ 2.512 [HF] [NaF] =10 0.40 ≈2.512 Final Ratio: NaF : HF = 2.51 : 1 NaF:HF=2.51:1 Question 4 Evaluate log ⁡ 3 81 log 3 81. Answer 1 Method 1: 81 = 3 4 81=3 4 : log ⁡ 3 81 = 4 log 3 81=4 Question 5 Evaluate log ⁡ 5 7 log 5 7. Answer 1 Using the change-of-base formula: log ⁡ 5 7 = log ⁡ 7 log ⁡ 5 log 5 7= log5 log7 Using approximations: log ⁡ ( 7 ) ≈ 0.8451 , log ⁡ ( 5 ) ≈ 0.6990 log(7)≈0.8451,log(5)≈0.6990 log ⁡ 5 7 = 0.8451 0.6990 ≈ 1.209 log 5 7= 0.6990 0.8451 ≈1.209