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

2024
Views:

146

Pages:

2
Author:

Reynaldo Wong

Question 1 Write in exponential form: log ⁡ 31 = 0 log31=0 Answer 1 A logarithmic equation of the form log ⁡ 𝑏 𝑥 = 𝑦 log b x=y can be converted to the exponential equation 𝑏 𝑦 = 𝑥 b y =x. Comparing log ⁡ 31 = 0 log31=0 to log ⁡ 𝑏 𝑥 = 𝑦 log b x=y, we have 𝑏 = 31 b=31, 𝑥 = 1 x=1, and 𝑦 = 0 y=0. Therefore, the exponential form is: 3 1 0 = 1 31 0 =1 Question 2 Convert ln ⁡ ( ln ⁡ 𝑥 ) = 2 ln(lnx)=2 to exponential form. Answer 1 Given: ln ⁡ ( ln ⁡ 𝑥 ) = 2 ln(lnx)=2 Step 1: Write in exponential form. ln ⁡ 𝑥 = 𝑒 2 lnx=e 2 Step 2: Write again in exponential form: 𝑥 = 𝑒 𝑒 2 x=e e 2 Result: 𝑥 ≈ 1618.178 x≈1618.178 Question 3 Logarithmic differentiation: Use logarithmic differentiation to evaluate 𝑓 ( 𝑥 ) = 𝑥 2 cos ⁡ 𝑥 f(x)=x 2cosx . Answer 1 Step 1: Logarithmic differentiation involves taking the natural logarithm of both sides of the function: ln ⁡ ( 𝑓 ( 𝑥 ) ) = ln ⁡ ( 𝑥 2 cos ⁡ 𝑥 ) ln(f(x))=ln(x 2cosx ) Using logarithmic properties: ln ⁡ ( 𝑓 ( 𝑥 ) ) = 2 cos ⁡ 𝑥 ln ⁡ 𝑥 ln(f(x))=2cosxlnx Step 2: Differentiate both sides with respect to 𝑥 x: 1 𝑓 ( 𝑥 ) 𝑑 𝑓 ( 𝑥 ) 𝑑 𝑥 = 2 cos ⁡ 𝑥 ⋅ 1 𝑥 + ln ⁡ 𝑥 ⋅ 𝑑 𝑑 𝑥 ( 2 cos ⁡ 𝑥 ) f(x) 1 dx df(x) =2cosx⋅ x 1 +lnx⋅ dx d (2cosx) Step 3: Simplify derivatives: 𝑑 𝑑 𝑥 ( 2 cos ⁡ 𝑥 ) = − 2 sin ⁡ 𝑥 dx d (2cosx)=−2sinx Substitute: 1 𝑓 ( 𝑥 ) 𝑑 𝑓 ( 𝑥 ) 𝑑 𝑥 = 2 cos ⁡ 𝑥 𝑥 − 2 ln ⁡ 𝑥 sin ⁡ 𝑥 f(x) 1 dx df(x) = x 2cosx −2lnxsinx Step 4: Multiply through by 𝑓 ( 𝑥 ) = 𝑥 2 cos ⁡ 𝑥 f(x)=x 2cosx : 𝑑 𝑓 ( 𝑥 ) 𝑑 𝑥 = 𝑥 2 cos ⁡ 𝑥 ( 2 cos ⁡ 𝑥 𝑥 − 2 ln ⁡ 𝑥 sin ⁡ 𝑥 ) dx df(x) =x 2cosx ( x 2cosx −2lnxsinx) Final Result: 𝑓 ′ ( 𝑥 ) = 𝑥 2 cos ⁡ 𝑥 ( 2 cos ⁡ 𝑥 𝑥 − 2 ln ⁡ 𝑥 sin ⁡ 𝑥 ) f ′ (x)=x 2cosx ( x 2cosx −2lnxsinx)