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[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 !!!!!
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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
mhanifa

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
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