University:
Shippensburg University of Pennsylvania
Course:
MAT 211 | Calculus I
Academic year:

2024
Views:

72

Pages:

4
Author:

Herbosambx

Question 1 Write the exponential equation as a logarithmic equation, or vice versa. 𝑒 − 2 = 0.1353 e −2 =0.1353 Answer Step 1: Given equation: 𝑒 − 2 = 0.1353 e −2 =0.1353 Take ln ⁡ ln on both sides: ln ⁡ ( 𝑒 − 2 ) = ln ⁡ ( 0.1353 ) ln(e −2 )=ln(0.1353) − 2 ln ⁡ ( 𝑒 ) = ln ⁡ ( 0.1353 ) −2ln(e)=ln(0.1353) − 2 = ln ⁡ ( 0.1353 ) −2=ln(0.1353) Question 2 Rewrite the Cartesian equation 𝑦 = 5 𝑥 2 y=5x 2 as a polar equation. Find 𝑟 ( 𝜃 ) = ? r(θ)=? Answer Step 1: Concept: In polar coordinates, 𝑥 = 𝑟 cos ⁡ 𝜃 , 𝑦 = 𝑟 sin ⁡ 𝜃 , and 𝑥 2 + 𝑦 2 = 𝑟 2 x=rcosθ,y=rsinθ,and x 2 +y 2 =r 2 Here, the objective is to express the Cartesian equation into polar equation. Step 2: Substitute 𝑟 sin ⁡ 𝜃 rsinθ for 𝑦 y and 𝑟 cos ⁡ 𝜃 rcosθ for 𝑥 x into the equation 𝑦 = 5 𝑥 2 y=5x 2 , and simplify as: 𝑟 sin ⁡ 𝜃 = 5 𝑟 2 cos ⁡ 2 𝜃 rsinθ=5r 2 cos 2 θ 𝑟 sin ⁡ 𝜃 = 5 𝑟 2 cos ⁡ 2 𝜃 rsinθ=5r 2 cos 2 θ 𝑟 = sin ⁡ 𝜃 5 cos ⁡ 2 𝜃 r= 5cos 2 θ sinθ Therefore, the polar equation is: 𝑟 ( 𝜃 ) = sin ⁡ 𝜃 5 cos ⁡ 2 𝜃 r(θ)= 5cos 2 θ sinθ Question 3 Use the definition of a logarithm to rewrite this equation as an exponential equation. log ⁡ 64 ( 8 ) = 1 2 log 64 (8)= 2 1 Answer Step 1: Given: log ⁡ 64 ( 8 ) = 1 2 log 64 (8)= 2 1 Step 2: Definition of logarithm: If log ⁡ 𝑏 ( 𝑎 ) = 𝑐 log b (a)=c, then 𝑏 𝑐 = 𝑎 b c =a, where 𝑏 b is called the base. Step 3: Using the above definition: log ⁡ 64 ( 8 ) = 1 2 can be written as 8 = ( 64 ) 1 2 log 64 (8)= 2 1 can be written as 8=(64) 2 1 Question 4 Decide whether the equation is an identity, a conditional equation, or a contradiction. Give the solution set. 12 ( 𝑥 − 2 ) = 2 ( 6 𝑥 + 1 ) − 26 12(x−2)=2(6x+1)−26 Answer 1: Step 1: Given: 12 ( 𝑥 − 2 ) = 2 ( 6 𝑥 + 1 ) − 26 12(x−2)=2(6x+1)−26 Step 2: Make the provided equation as straightforward as possible. Utilize the distribution property: 12 ( 𝑥 − 2 ) = 2 ( 6 𝑥 ) + 2 ( 1 ) − 26 12(x−2)=2(6x)+2(1)−26 12 𝑥 − 24 = 12 𝑥 − 24 12x−24=12x−24 Step 3: Subtract 12 𝑥 12x on both sides: 12 𝑥 − 24 − 12 𝑥 = 12 𝑥 − 24 − 12 𝑥 12x−24−12x=12x−24−12x − 24 = − 24 −24=−24 The equation is an identity solution. Final Answer: The solution set of the given equation is the set of all real numbers. Question 5 Write the equation of a line passing through the points ( − 5 , − 5 ) (−5,−5) and − 1 , − 7 −1,−7. Write the equation in slope-intercept form. Answer Step 1: Slope of the line passing through ( 𝑥 1 , 𝑦 1 ) (x 1 ,y 1 ) and ( 𝑥 2 , 𝑦 2 ) (x 2 ,y 2 ) is given by: 𝑚 = 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1 m= x 2 −x 1 y 2 −y 1 Given: A line is passing through ( − 5 , − 5 ) (−5,−5) and ( − 1 , − 7 ) (−1,−7). 𝑚 = − 7 − ( − 5 ) − 1 − ( − 5 ) m= −1−(−5) −7−(−5) 𝑚 = − 2 4 = − 1 2 m= 4 −2 =− 2 1 The equation of the line is 𝑦 = 𝑚 𝑥 + 𝑐 y=mx+c. Substitute the values of 𝑚 m and one point: Plug in 𝑥 = − 5 x=−5 and 𝑦 = − 5 y=−5 in 𝑦 = − 1 2 𝑥 + 𝑐 y=− 2 1 x+c: − 5 = − 1 2 ( − 5 ) + 𝑐 −5=− 2 1 (−5)+c − 5 = 5 2 + 𝑐 −5= 2 5 +c 𝑐 = − 5 − 5 2 = − 15 2 c=−5− 2 5 =− 2 15 Final Equation: 𝑦 = − 1 2 𝑥 − 15 2 y=− 2 1 x− 2 15