Answered

Connect with knowledgeable experts and enthusiasts on IDNLearn.com. Receive prompt and accurate responses to your questions from our community of knowledgeable professionals ready to assist you at any time.


[tex]Evaluate \: the \: following \\ (1) \: log1000 \: \\ (2) ( \frac{128}{625} ) \\ (3) log {x}^{2} {y}^{3} {z}^{4} \\ (4) log \frac{ {p}^{2} {q}^{3} }{r} \\ (5) log \sqrt{ \frac{ {x}^{3} }{ {y}^{2} } } [/tex]
[tex]If \: {x}^{2} + {y}^{2} = 25xy. \\ Then \: prove \: that \: \\ 2 log(x + y) = \\ 3 log3 + logx + logy. \: [/tex]
[tex] \: \: \: \: \: \: \: \: \: \: \: \: \:I \: need \: the \: answer \\ \: \: \: \: \: \: \: \: \: \: \: \: Plz \: fastly \: [/tex]
I need ans !!!!!
​

Sagot :

Step-by-step explanation:

  1. log1000= log 10³= 3 log10 =3
  2. log(128/625)= 7 log 2+ 4 log 5
  3. log x²y³z⁴= 2 logx + 3 log y + 4 log z
  4. log p²q³/r= 2 log p +3 log q - log r
  5. log√(x³/y²)=3/2[ log (x)] - log y

x²+y²=25xy

(x+y)²-2xy=25xy

(x+y)²= 2xy +25 xy

=27xy

Take log on both sides

2 log(x+y) =log 27 + log x + log y

=log 3³+ log x + log y

2 log(x+y)=3 log 3 + log x + log y

View image Аноним
Mhanifaidn

Answer:

(1)

  • log 1000 = log 10³ = 3 log 10 = 3

(2) It should be with log? If yes ignore log x and consider the right side

  • (128/625) = x
  • log x = log (128/625)
  • log x = log 128 - log 625
  • log x = 7 log 2 - 4 log 5

(3)

  • log (x²y³z⁴) = log x² + log y³ + log z⁴ = 2 log x + 3 log y + 4 log z

(4)

  • log (p²q³/r) = log p² + log q³ - log r = 2 log p + 3 log q - log r

(5)

  • log [tex]\sqrt{\frac{x^3}{y2} }[/tex] = 1/2 log [tex]x^3y^{-2}[/tex] = 3/2 log x - log y

(6)

  • x² + y² = 25xy
  • x² + 2xy + y² = 27xy
  • (x + y)² = 27xy
  • log (x + y)² = log (27xy)
  • 2 log (x + y) = log 3³+ log x + log y
  • 2 log (x + y) = 3 log 3 + log x + log y
  • Proved
Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. Thank you for choosing IDNLearn.com. We’re here to provide reliable answers, so please visit us again for more solutions.