Logarithms #2 - Q&A

Question 1 Use the properties of logarithms to expand the following expression as much as possible. Simplify any numerical expressions that can be evaluated without a calculator. log ⁡ 4 ( 48 𝑥 + 16 𝑦 ) log 4 ​ (48x+16y) Answer 1 Step 1: Given The following expression: log ⁡ 4 ( 48 𝑥 + 16 𝑦 ) log 4 ​ (48x+16y) Step 2: To determine Expand the following expression using the properties of logarithms. Let log ⁡ 4 ( 48 𝑥 + 16 𝑦 ) = log ⁡ ( 48 𝑥 + 16 𝑦 ) log ⁡ ( 4 ) log 4 ​ (48x+16y)=log log(4) (48x+16y) ​ (by the Change of Base formula). Expanding the above equation, we get: log ⁡ 4 ( 48 𝑥 + 16 𝑦 ) = log ⁡ ( 48 ) + log ⁡ ( 3 𝑥 + 𝑦 ) log ⁡ ( 4 ) log 4 ​ (48x+16y)= log(4) log(48)+log(3x+y) ​ = log ⁡ ( 16 ) + log ⁡ ( 3 𝑥 + 𝑦 ) log ⁡ ( 4 ) = log(4) log(16)+log(3x+y) ​ = log ⁡ ( 4 ) ( 𝑎 ) −   𝑎 𝑛 𝑑 𝑖 𝑓 𝑆 𝑖 𝑚 𝑝 𝑙 𝑖 𝑓 𝑖 𝑒 𝑑 𝑆 𝑖 𝑚 𝑖 𝑙 𝑎 𝑟 𝑙 𝑦 𝐹 𝑟 𝑜 𝑚 ( 𝑒 𝑥 𝑝 𝑎 𝑛 𝑠 𝑖 𝑜 𝑛 𝑆 𝑖 𝑚 𝑝 𝑙 𝑖 𝑓 𝑖 𝑒 𝑠 = 𝑆 𝑈 𝑀 𝑇 𝑂 𝑊 𝐴 𝑅 𝐷 𝑆 ) + − − > ( ( 𝑟 𝑒 𝑤 𝑟 𝑖 𝑡 𝑒 − 𝑏 𝑦 − 𝑙 𝑜 𝑔 𝑠 𝑖 𝑚 𝑝 𝑙 𝑖 𝑓 𝑖 𝑒 𝑑 − 𝑓 𝑜 𝑟 𝑚 𝐸 𝑥 𝑝 𝑎 𝑛 𝑑 ∗ ∗ 𝑆 𝑇 𝐸 𝑃 𝐵 𝑎 𝑐 𝑘 = 𝑚 𝑎 𝑡 ℎ 𝑒 𝑛 𝑑 𝑠 𝑡 𝑜 +   ( ( 𝑟 𝑒 𝑤 𝑟 𝑖 𝑡 𝑒 − 𝑠 𝑢 𝑚 𝑚 𝑎 𝑟 𝑦 + 𝑟 𝑢 𝑙 𝑒 𝑠 ∗ ∗ = − log(4)(a) ​  andifSimplifiedSimilarlyFrom(expansionSimplifies=SUMTOWARDS)+−−>((rewrite−by−logsimplified−formExpand∗∗STEPBack=mathendsto+((rewrite−summary+rules∗∗ It seems my transcription got mixed up earlier. Let me correct and clearly state the extracted answers from the image in a well-formatted way. Question 1 Use the properties of logarithms to expand the following expression as much as possible. Simplify any numerical expressions that can be evaluated without a calculator. log ⁡ 4 ( 48 𝑥 + 16 𝑦 ) log 4 ​ (48x+16y) Answer 1 Step 1: Expand using the properties of logarithms. Given: log ⁡ 4 ( 48 𝑥 + 16 𝑦 ) = log ⁡ ( 48 𝑥 + 16 𝑦 ) log ⁡ ( 4 ) (Change of Base formula) log 4 ​ (48x+16y)= log(4) log(48x+16y) ​ (Change of Base formula) Expanding log ⁡ ( 48 𝑥 + 16 𝑦 ) log(48x+16y): log ⁡ ( 48 𝑥 + 16 𝑦 ) = log ⁡ ( 16 ) + log ⁡ ( 3 𝑥 + 𝑦 ) log(48x+16y)=log(16)+log(3x+y) Substitute this back: log ⁡ 4 ( 48 𝑥 + 16 𝑦 ) = log ⁡ ( 16 ) + log ⁡ ( 3 𝑥 + 𝑦 ) log ⁡ ( 4 ) Substitute this back:log 4 ​ (48x+16y)= log(4) log(16)+log(3x+y) ​ Simplify log ⁡ ( 16 ) log(16): log ⁡ ( 16 ) = log ⁡ ( 4 2 ) = 2 log ⁡ ( 4 ) log(16)=log(4 2 )=2log(4) Thus: log ⁡ 4 ( 48 𝑥 + 16 𝑦 ) = 2 log ⁡ ( 4 ) + log ⁡ ( 3 𝑥 + 𝑦 ) log ⁡ ( 4 ) log 4 ​ (48x+16y)= log(4) 2log(4)+log(3x+y) ​ Split the fraction: = 2 + log ⁡ ( 3 𝑥 + 𝑦 ) log ⁡ ( 4 ) =2+ log(4) log(3x+y) ​ Final answer: log ⁡ 4 ( 48 𝑥 + 16 𝑦 ) = 2 + log ⁡ 4 ( 3 𝑥 + 𝑦 ) log 4 ​ (48x+16y)=2+log 4 ​ (3x+y) Question 2 Rewrite the logarithm as a ratio of common logarithms and natural ones. log ⁡ 2 ( 3 7 ) log 2 ​ ( 7 3 ​ ) Answer 1 Using the Change of Base formula: log ⁡ 2 ( 3 7 ) = log ⁡ ( 3 7 ) log ⁡ ( 2 ) log 2 ​ ( 7 3 ​ )= log(2) log( 7 3 ​ ) ​ Further simplify log ⁡ ( 3 7 ) log( 7 3 ​ ) using logarithmic properties: log ⁡ ( 3 7 ) = log ⁡ ( 3 ) − log ⁡ ( 7 ) log( 7 3 ​ )=log(3)−log(7) Thus: log ⁡ 2 ( 3 7 ) = log ⁡ ( 3 ) − log ⁡ ( 7 ) log ⁡ ( 2 ) log 2 ​ ( 7 3 ​ )= log(2) log(3)−log(7) ​ Alternatively, using natural logarithms: log ⁡ 2 ( 3 7 ) = ln ⁡ ( 3 ) − ln ⁡ ( 7 ) ln ⁡ ( 2 ) log 2 ​ ( 7 3 ​ )= ln(2) ln(3)−ln(7) ​ Question 3 The One-to-One Property of natural logarithms states that if ln ⁡ 𝑥 = ln ⁡ 𝑦 lnx=lny, then ________. Answer 1 If ln ⁡ 𝑥 = ln ⁡ 𝑦 lnx=lny, then: 𝑥 = 𝑦 x=y Explanation: The natural logarithm is a one-to-one function, which means it has a unique output for every input. If ln ⁡ 𝑥 = ln ⁡ 𝑦 lnx=lny, their inverses (exponential values) must also be equal: 𝑒 ln ⁡ 𝑥 = 𝑒 ln ⁡ 𝑦 ⇒ 𝑥 = 𝑦 e lnx =e lny ⇒x=y Question 4 How do you simplify 𝑒 ln ⁡ 𝑥 e lnx ? Answer 1 Solution: By definition of the natural logarithm: 𝑒 ln ⁡ 𝑥 = 𝑥 e lnx =x Explanation: The natural logarithm ln ⁡ 𝑥 lnx is the power to which 𝑒 e must be raised to obtain 𝑥 x. Thus, raising 𝑒 e to the power of ln ⁡ 𝑥 lnx simply gives 𝑥 x. Question 5 How do you solve ln ⁡ 𝑥 + ln ⁡ ( 𝑥 − 1 ) = 1 lnx+ln(x−1)=1? Answer 1 Step 1: Combine the logarithms. Using ln ⁡ 𝑎 + ln ⁡ 𝑏 = ln ⁡ ( 𝑎 𝑏 ) lna+lnb=ln(ab): ln ⁡ 𝑥 + ln ⁡ ( 𝑥 − 1 ) = ln ⁡ [ 𝑥 ( 𝑥 − 1 ) ] lnx+ln(x−1)=ln[x(x−1)] The equation becomes: ln ⁡ [ 𝑥 ( 𝑥 − 1 ) ] = 1 ln[x(x−1)]=1 Step 2: Exponentiate both sides to eliminate the logarithm. 𝑒 ln ⁡ [ 𝑥 ( 𝑥 − 1 ) ] = 𝑒 1 e ln[x(x−1)] =e 1 𝑥 ( 𝑥 − 1 ) = 𝑒 x(x−1)=e Step 3: Expand and rewrite as a quadratic equation. 𝑥 2 − 𝑥 = 𝑒 x 2 −x=e 𝑥 2 − 𝑥 − 𝑒 = 0 x 2 −x−e=0 Step 4: Solve the quadratic equation using the quadratic formula: 𝑥 = − 𝑏 ± 𝑏 2 − 4 𝑎 𝑐 2 𝑎 x= 2a −b± b 2 −4ac ​ ​ Here: 𝑎 = 1 a=1, 𝑏 = − 1 b=−1, 𝑐 = − 𝑒 c=−e. Substitute: 𝑥 = − ( − 1 ) ± ( − 1 ) 2 − 4 ( 1 ) ( − 𝑒 ) 2 ( 1 ) x= 2(1) −(−1)± (−1) 2 −4(1)(−e) ​ ​ 𝑥 = 1 ± 1 + 4 𝑒 2 x= 2 1± 1+4e ​ ​ Step 5: Select the valid solution. Since 𝑥 > 0 x>0: 𝑥 = 1 + 1 + 4 𝑒 2 x= 2 1+ 1+4e ​ ​ Final Answer: 𝑥 = 1 + 1 + 4 𝑒 2 x= 2 1+ 1+4e