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

2024
Views:

39

Pages:

3
Author:

Reynaldo Wong

Question 1 Which equation is equivalent to log ⁡ 𝑦 = 2 𝑥 + 1 logy=2x+1? (A) y = 10^{(100)}^x (B) 𝑦 = 1 0 2 𝑥 y=10 2x (C) 𝑦 = 𝑒 2 𝑥 + 1 y=e 2x +1 (D) 𝑦 = 𝑒 2 y=e 2 Answer 1 Exponentiate both sides of the equation and simplify: log ⁡ 𝑦 = 2 𝑥 + 1 logy=2x+1 1 0 log ⁡ 𝑦 = 1 0 2 𝑥 + 1 10 logy =10 2x+1 𝑦 = 1 0 2 𝑥 ⋅ 1 0 1 y=10 2x ⋅10 1 𝑦 = 1 0 2 𝑥 ⋅ 10 y=10 2x ⋅10 𝑦 = 1 0 ( 100 ) 𝑥 y=10 (100) x Equation A Question 2 What is ln ⁡ ( 𝑒 𝑥 ) ln(e x )? Answer 1 Explanation: You are seeking a number that is the exponent of ln ⁡ ln base, which provides us with the integrand, 𝑒 𝑥 e x . So: The base of ln ⁡ ln is 𝑒 e. The exponent of this base must be this number in order to obtain 𝑒 𝑥 e x , exactly 𝑥 x. ln ⁡ ( 𝑒 𝑥 ) = log ⁡ 𝑒 ( 𝑒 𝑥 ) = 𝑥 ln(e x )=log e (e x )=x Answer 2 Explanation: ln ⁡ ( 𝑒 𝑥 ) = 𝑥 because log ⁡ 𝑎 ( 𝑎 𝑥 ) = 𝑥 ln(e x )=xbecauselog a (a x )=x Answer 3 ln ⁡ ( 𝑒 𝑥 ) = 𝑥 ln(e x )=x Here, ln ⁡ ln denotes the natural logarithm, 𝑒 e represents the base of the natural logarithm, and 𝑥 x is the variable. Question 3 What is the derivative of ln ⁡ ( 3 𝑥 ) ln(3x)? Answer 1 ln ⁡ ( 3 𝑥 ) = 𝑦 ln(3x)=y 𝑒 𝑦 = 3 𝑥 e y =3x Use implicit differentiation: 𝑑 𝑦 𝑑 𝑦 = 𝑑 𝑦 𝑑 𝑥 dy dy = dx dy 𝑒 𝑦 = 3 𝑥 transforms into: e y =3xtransforms into: 𝑑 𝑦 / 𝑑 𝑥 = 3 dy/dx=3 Sorry, it seems there was a mistake. Let me retype the continuation from Question 3 Answer 1: What is the derivative of ln ⁡ ( 3 𝑥 ) ln(3x)? Answer 1 (Continued): Using the chain rule: 𝑑 𝑑 𝑥 [ ln ⁡ ( 3 𝑥 ) ] = 1 3 𝑥 ⋅ 𝑑 𝑑 𝑥 ( 3 𝑥 ) = 1 3 𝑥 ⋅ 3 = 1 𝑥 dx d [ln(3x)]= 3x 1 ⋅ dx d (3x)= 3x 1 ⋅3= x 1 Alternatively, rewrite ln ⁡ ( 3 𝑥 ) ln(3x): ln ⁡ ( 3 𝑥 ) = ln ⁡ ( 3 ) + ln ⁡ ( 𝑥 ) ln(3x)=ln(3)+ln(x) Taking the derivative: 𝑑 𝑑 𝑥 [ ln ⁡ ( 3 ) + ln ⁡ ( 𝑥 ) ] = 0 + 1 𝑥 = 1 𝑥 dx d [ln(3)+ln(x)]=0+ x 1 = x 1 Question 4 Use properties of logarithms to condense the logarithmic expression log ⁡ 5 + log ⁡ 2 log5+log2. Write the expression as a single logarithm whose coefficient is 1. Where possible, evaluate logarithmic expressions without using a calculator. Answer 1 We have to condense the logarithmic expression: log ⁡ ( 5 ) + log ⁡ ( 2 ) log(5)+log(2) Using the property of logarithms: log ⁡ ( 𝑎 ) + log ⁡ ( 𝑏 ) = log ⁡ ( 𝑎 𝑏 ) log(a)+log(b)=log(ab) Apply this property: log ⁡ ( 5 ) + log ⁡ ( 2 ) = log ⁡ ( 5 ⋅ 2 ) = log ⁡ ( 10 ) log(5)+log(2)=log(5⋅2)=log(10) If the base of the logarithm is 10: log ⁡ 10 ( 10 ) = 1 log 10 (10)=1 Thus, the condensed expression is: 1 1 Question 5 Solve: ln ⁡ ( 𝑥 ) = ln ⁡ ( 𝑥 + 6 ) − ln ⁡ ( 𝑥 − 4 ) ln(x)=ln(x+6)−ln(x−4) Answer 1 Step 1: Using the property of logarithms ln ⁡ ( 𝑎 ) − ln ⁡ ( 𝑏 ) = ln ⁡ ( 𝑎 𝑏 ) ln(a)−ln(b)=ln( b a ), simplify the equation: ln ⁡ ( 𝑥 ) = ln ⁡ ( 𝑥 + 6 𝑥 − 4 ) ln(x)=ln( x−4 x+6 ) Step 2: Since ln ⁡ ( 𝑎 ) = ln ⁡ ( 𝑏 ) ln(a)=ln(b) implies 𝑎 = 𝑏 a=b: 𝑥 = 𝑥 + 6 𝑥 − 4 x= x−4 x+6 Step 3: Eliminate the fraction by multiplying through by ( 𝑥 − 4 ) (x−4): 𝑥 ( 𝑥 − 4 ) = 𝑥 + 6 x(x−4)=x+6 𝑥 2 − 4 𝑥 = 𝑥 + 6 x 2 −4x=x+6 𝑥 2 − 5 𝑥 − 6 = 0 x 2 −5x−6=0 Step 4: Factorize the quadratic: 𝑥 2 − 5 𝑥 − 6 = ( 𝑥 − 6 ) ( 𝑥 + 1 ) = 0 x 2 −5x−6=(x−6)(x+1)=0 𝑥 = 6 , 𝑥 = − 1 x=6,x=−1 Step 5: Since ln ⁡ ( 𝑥 ) ln(x) is undefined for non-positive 𝑥 x, discard 𝑥 = − 1 x=−1. Thus, the solution is: 𝑥 = 6 x=6