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

2024
Views:

99

Pages:

3
Author:

Reynaldo Wong

Question 1 How do you solve ln ⁡ 𝑥 + ln ⁡ ( 𝑥 − 1 ) = 1 lnx+ln(x−1)=1? Answer 1 Making use of the logarithm rules: ln ⁡ ( 𝑥 ) + ln ⁡ ( 𝑥 − 1 ) = ln ⁡ ( 𝑥 ⋅ ( 𝑥 − 1 ) ) = ln ⁡ ( 𝑥 2 − 𝑥 ) ln(x)+ln(x−1)=ln(x⋅(x−1))=ln(x 2 −x) The equation becomes: ln ⁡ ( 𝑥 2 − 𝑥 ) = 1 ln(x 2 −x)=1 Exponentiate both sides: 𝑒 ln ⁡ ( 𝑥 2 − 𝑥 ) = 𝑒 1 e ln(x 2 −x) =e 1 𝑥 2 − 𝑥 = 𝑒 x 2 −x=e Solve for 𝑥 x: Step 1: Complete the square: 𝑥 2 − 𝑥 = 𝑒 x 2 −x=e 𝑥 2 − 𝑥 + 1 4 = 𝑒 + 1 4 x 2 −x+ 4 1 =e+ 4 1 ( 𝑥 − 1 2 ) 2 = 𝑒 + 1 4 (x− 2 1 ) 2 =e+ 4 1 Step 2: Solve for 𝑥 x: 𝑥 − 1 2 = ± 𝑒 + 1 4 x− 2 1 =± e+ 4 1 𝑥 = 1 2 ± 𝑒 + 1 4 x= 2 1 ± e+ 4 1 Step 3: Exclude negative values since 𝑥 > 0 x>0: 𝑥 = 1 2 + 𝑒 + 1 4 x= 2 1 + e+ 4 1 Final Answer: 𝑥 = 1 + 1 + 4 𝑒 2 x= 2 1+ 1+4e Question 2 Solve: log ⁡ 6 𝑥 = 0.5 log ⁡ 636 log6x=0.5log636 Answer 1 Simplify using logarithmic properties: log ⁡ ( 6 𝑥 ) = log ⁡ ( 63 6 0.5 ) log(6x)=log(636 0.5 ) log ⁡ ( 6 𝑥 ) = log ⁡ ( 636 ) log(6x)=log( 636 ) Simplify further: 6 𝑥 = 636 6x= 636 Divide both sides by 6: 𝑥 = 6 x=6 Question 3 Simplify log ⁡ 3 ( 1 9 ) log 3 ( 9 1 ). Answer 1 Rewrite 1 9 9 1 as a power of 3: log ⁡ 3 ( 1 9 ) = log ⁡ 3 ( 3 − 2 ) log 3 ( 9 1 )=log 3 (3 −2 ) Use the property log ⁡ 𝑏 ( 𝑎 𝑐 ) = 𝑐 log ⁡ 𝑏 ( 𝑎 ) log b (a c )=clog b (a): log ⁡ 3 ( 3 − 2 ) = − 2 log 3 (3 −2 )=−2 Final Answer: − 2 −2 Question 4 Write ln ⁡ ( 𝑥 − 2 ) = 5 ln(x−2)=5 in exponential form. Answer 1 By definition of natural logarithm: ln ⁡ ( 𝑥 − 2 ) = 5 ⟹ 𝑒 5 = 𝑥 − 2 ln(x−2)=5⟹e 5 =x−2 Rearranging: 𝑥 = 𝑒 5 + 2 x=e 5 +2 Question 5 Use properties of logarithms to condense the logarithmic expression. Write the expression as a single logarithm whose coefficient is 1: 1 3 ( log ⁡ 𝑢 ( 𝑥 ) − log ⁡ 𝑢 ( 𝑦 ) ) 3 1 (log u (x)−log u (y)) Answer 1 Using the properties of logarithms: log ⁡ 𝑢 ( 𝑎 ) − log ⁡ 𝑢 ( 𝑏 ) = log ⁡ 𝑢 ( 𝑎 𝑏 ) log u (a)−log u (b)=log u ( b a ) 1 3 ( log ⁡ 𝑢 ( 𝑥 ) − log ⁡ 𝑢 ( 𝑦 ) ) = 1 3 log ⁡ 𝑢 ( 𝑥 𝑦 ) 3 1 (log u (x)−log u (y))= 3 1 log u ( y x ) Apply the power rule 𝑐 ⋅ log ⁡ 𝑏 ( 𝑎 ) = log ⁡ 𝑏 ( 𝑎 𝑐 ) c⋅log b (a)=log b (a c ): 1 3 log ⁡ 𝑢 ( 𝑥 𝑦 ) = log ⁡ 𝑢 ( ( 𝑥 𝑦 ) 1 3 ) 3 1 log u ( y x )=log u (( y x ) 3 1 ) Final Answer: log ⁡ 𝑢 ( 𝑥 𝑦 3 ) log u ( 3 y x )