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

2024
Views:

317

Pages:

2
Author:

Reynaldo Wong

Question 1 Using the change-of-base formula, evaluate the logarithm. Round your result to three decimal places: log ⁡ 36 ( 25 ) log 36 (25) Answer 1 Using the change-of-base formula: log ⁡ 36 ( 25 ) = log ⁡ 10 ( 25 ) log ⁡ 10 ( 36 ) log 36 (25)= log 10 (36) log 10 (25) Using a calculator: log ⁡ 10 ( 25 ) ≈ 1.398 , log ⁡ 10 ( 36 ) ≈ 1.556 log 10 (25)≈1.398,log 10 (36)≈1.556 log ⁡ 36 ( 25 ) = 1.398 1.556 ≈ 0.898 log 36 (25)= 1.556 1.398 ≈0.898 Question 2 Find the sum, difference, or product of logarithms as a single logarithm, if possible, using the properties of logarithms: log ⁡ 𝑎 13 − log ⁡ 𝑎 4 log a 13−log a 4 Answer 1 The logarithm is: log ⁡ 𝑎 13 − log ⁡ 𝑎 4 log a 13−log a 4 Using the quotient rule of logarithms: log ⁡ 𝑎 ( 𝑚 ) − log ⁡ 𝑎 ( 𝑛 ) = log ⁡ 𝑎 ( 𝑚 𝑛 ) log a (m)−log a (n)=log a ( n m ) Apply the rule: log ⁡ 𝑎 13 − log ⁡ 𝑎 4 = log ⁡ 𝑎 ( 13 4 ) log a 13−log a 4=log a ( 4 13 ) Final Answer: log ⁡ 𝑎 ( 13 4 ) log a ( 4 13 ) Question 3 Write each expression as a single natural logarithm: 2 ln ⁡ 8 − 3 ln ⁡ 4 2ln8−3ln4 Answer 1 Using the power rule of logarithms: 𝑛 ln ⁡ ( 𝑚 ) = ln ⁡ ( 𝑚 𝑛 ) nln(m)=ln(m n ) 2 ln ⁡ ( 8 ) − 3 ln ⁡ ( 4 ) = ln ⁡ ( 8 2 ) − ln ⁡ ( 4 3 ) 2ln(8)−3ln(4)=ln(8 2 )−ln(4 3 ) Simplify the powers: ln ⁡ ( 64 ) − ln ⁡ ( 64 ) ln(64)−ln(64) Using the quotient rule: ln ⁡ ( 𝑚 ) − ln ⁡ ( 𝑛 ) = ln ⁡ ( 𝑚 𝑛 ) ln(m)−ln(n)=ln( n m ) ln ⁡ ( 64 ) − ln ⁡ ( 64 ) = ln ⁡ ( 64 64 ) = ln ⁡ ( 1 ) ln(64)−ln(64)=ln( 64 64 )=ln(1) Final Answer: ln ⁡ ( 1 ) = 0 ln(1)=0 Question 4 Express the given quantity as a single logarithm: 1 3 ln ⁡ ( 𝑥 + 2 ) + 1 2 ln ⁡ ( 𝑥 − 2 𝑥 2 + 3 𝑥 + 2 ) 3 1 ln(x+2)+ 2 1 ln( x 2 +3x+2 x−2 ) Answer 1 Expand step by step: Step 1: Combine the exponents. ln ⁡ ( 𝑥 + 2 ) 1 / 3 + ln ⁡ ( 𝑥 − 2 𝑥 2 + 3 𝑥 + 2 ) 1 / 2 ln(x+2) 1/3 +ln( x 2 +3x+2 x−2 ) 1/2 Step 2: Use the product rule of logarithms: ln ⁡ ( 𝑎 ) + ln ⁡ ( 𝑏 ) = ln ⁡ ( 𝑎 ⋅ 𝑏 ) ln(a)+ln(b)=ln(a⋅b) ln ⁡ ( ( 𝑥 + 2 ) 1 / 3 ⋅ ( 𝑥 − 2 𝑥 2 + 3 𝑥 + 2 ) 1 / 2 ) ln((x+2) 1/3 ⋅( x 2 +3x+2 x−2 ) 1/2 ) Step 3: Simplify: Factor 𝑥 2 + 3 𝑥 + 2 = ( 𝑥 + 2 ) ( 𝑥 + 1 ) x 2 +3x+2=(x+2)(x+1): ln ⁡ ( ( 𝑥 + 2 ) 1 / 3 ⋅ ( 𝑥 − 2 ) 1 / 2 ( 𝑥 + 2 ) 1 / 2 ( 𝑥 + 1 ) 1 / 2 ) ln( (x+2) 1/2 (x+1) 1/2 (x+2) 1/3 ⋅(x−2) 1/2 ) Step 4: Cancel ( 𝑥 + 2 ) 1 / 3 (x+2) 1/3 with part of ( 𝑥 + 2 ) 1 / 2 (x+2) 1/2 : ln ⁡ ( ( 𝑥 + 2 ) − 1 / 6 ⋅ ( 𝑥 − 2 ) 1 / 2 ( 𝑥 + 1 ) 1 / 2 ) ln( (x+1) 1/2 (x+2) −1/6 ⋅(x−2) 1/2 ) Final Answer: ln ⁡ ( ( 𝑥 − 2 ) 1 / 2 ( 𝑥 + 2 ) 1 / 6 ( 𝑥 + 1 ) 1 / 2 ) ln( (x+2) 1/6 (x+1) 1/2 (x−2) 1/2 ) Question 5 Describe the product rule for logarithms and give an example. Answer 1 The product rule for logarithms states: log ⁡ 𝑏 ( 𝑚 ⋅ 𝑛 ) = log ⁡ 𝑏 ( 𝑚 ) + log ⁡ 𝑏 ( 𝑛 ) log b (m⋅n)=log b (m)+log b (n) Example: Let 𝑚 = 2 m=2, 𝑛 = 10 n=10, 𝑏 = 10 b=10: log ⁡ 10 ( 2 ⋅ 10 ) = log ⁡ 10 ( 2 ) + log ⁡ 10 ( 10 ) log 10 (2⋅10)=log 10 (2)+log 10 (10) Simplify: log ⁡ 10 ( 20 ) = log ⁡ 10 ( 2 ) + 1 log 10 (20)=log 10 (2)+1 Conclusion: The product rule allows you to separate the logarithm of a product into a sum of individual logarithms.