Use logarithm properties to expand the given expressions.
Apply the power rule: ln ( x n ) = n ln x .
Apply the product rule: ln ( x y ) = ln x + ln y .
Substitute the given values and simplify: (a) − 26 , (b) 7 .
Explanation
Problem Setup and Strategy We are given that ln a = 2 , ln b = 3 , and ln c = 5 . We need to evaluate two expressions involving logarithms and the variables a , b , and c . We will use the properties of logarithms to simplify the expressions and then substitute the given values.
Evaluating Part (a) (a) We want to evaluate ln ( b 4 c 4 a 3 ) . Using the properties of logarithms, we can rewrite this as: ln ( b 4 c 4 a 3 ) = ln ( a 3 ) − ln ( b 4 c 4 ) Using the product rule for logarithms, we have: ln ( b 4 c 4 ) = ln ( b 4 ) + ln ( c 4 ) So, ln ( b 4 c 4 a 3 ) = ln ( a 3 ) − ( ln ( b 4 ) + ln ( c 4 )) Using the power rule for logarithms, we have ln ( x n ) = n ln x . Thus, ln ( a 3 ) = 3 ln a , ln ( b 4 ) = 4 ln b , ln ( c 4 ) = 4 ln c Substituting these into our expression, we get: ln ( b 4 c 4 a 3 ) = 3 ln a − ( 4 ln b + 4 ln c ) Now, we substitute the given values ln a = 2 , ln b = 3 , and ln c = 5 :
ln ( b 4 c 4 a 3 ) = 3 ( 2 ) − ( 4 ( 3 ) + 4 ( 5 )) = 6 − ( 12 + 20 ) = 6 − 32 = − 26
Evaluating Part (b) (b) We want to evaluate ln a 3 b 1 c 1 . We can rewrite the square root as a power of 2 1 :
ln a 3 b 1 c 1 = ln ( a 3 b 1 c 1 ) 2 1 Using the power rule for logarithms, we have: ln ( a 3 b 1 c 1 ) 2 1 = 2 1 ln ( a 3 b 1 c 1 ) Using the product rule for logarithms, we have: ln ( a 3 b 1 c 1 ) = ln ( a 3 ) + ln ( b 1 ) + ln ( c 1 ) Using the power rule for logarithms again, we have: ln ( a 3 ) = 3 ln a , ln ( b 1 ) = 1 ln b , ln ( c 1 ) = 1 ln c Substituting these into our expression, we get: 2 1 ln ( a 3 b 1 c 1 ) = 2 1 ( 3 ln a + 1 ln b + 1 ln c ) Now, we substitute the given values ln a = 2 , ln b = 3 , and ln c = 5 :
2 1 ( 3 ln a + 1 ln b + 1 ln c ) = 2 1 ( 3 ( 2 ) + 1 ( 3 ) + 1 ( 5 )) = 2 1 ( 6 + 3 + 5 ) = 2 1 ( 14 ) = 7
Final Answer (a) ln ( b 4 c 4 a 3 ) = − 26 (b) ln a 3 b 1 c 1 = 7
Examples
Logarithms are incredibly useful in many fields, including finance. For example, continuously compounded interest uses the natural logarithm. If you invest an amount P at an annual interest rate r , compounded continuously, the amount A you'll have after t years is given by A = P e r t . Logarithms can help solve for any of these variables. Understanding logarithmic properties allows for efficient calculation and manipulation of financial models, helping investors make informed decisions.