Expand the following logarithm completely

  ((3 ln 3 +  3 ln x) +  0.5 (ln (y^{2}-1) −

    { (2 (ln y) [ ln (x-1) ]  ;  

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Step-by-step explanation:

Given problem:  Expand the following logarithm completely:

→   ] ;

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To begin:  Simplify:

    →     ]  ;

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First, within the "numerator" ,  rewrite:  " √(y² - 1) " ;  as:

         "   " ;

→  Note the following property of exponents:

            ;

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Then rewrite the expression as:

Note the "quotient rule" property of the "natural logarithm (ln)" ;

          ln (a/b)  = ln a − ln b ;

So; we have:

      ;

 ↔  

 -

    ln ;

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    ln a = ln    ;

    ln b = ln ;

Simplify:  ln a − ln b  ;

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First, simplify:  ln a ;  

     ln a = ln    ;

→  Note the "product rule" for the "natural logarithm" {" ln " }:

  →    ln(a * b) = ln(a) + ln(b) ;

  →    ln (  * ) ;

             in which:

                a = ;

                b = ;

→  ln (a * b) =

    (ln a  +  ln b ) =

      =  [ ln ()] + [ ln [ = ? ;

First, simplify:  "ln a " ;

→ ln a = ln = ? ;

→  Note the "product rule" for the "natural logarithm" {" ln " }:

  →    ln(a * b) = ln(a) + ln(b) ;

  →    ln {27 *x^{3}) = ln 27 + ln x³  ;  in which:  a = 27 ; b = x³  ;

  →    ln(a * b) = ln(a) + ln(b) = ln 27 + ln (x³)  = ?  ;

Note:   We have "x³ " ;  Note the number:  "27" ;  the ∛27 = 3 ; a whole number integer; so rewrite "ln 27" as:  "ln (3³)"

  →    ln(a * b) = ln(a) + ln(b) = ln 27 + ln (x³) = ln (3³) + ln (x³) ;

                                                                     =  3 ln 3 + 3 ln x .  

So;  "ln a " from above:  →  ln(a * b) =  (ln a + ln b) ;

      =  [ ln ()] + [ ln [  = ? ;

→ Rewrite:  (3 ln 3 +  3 ln x) + [ln [ = ? ;

Now, find simplify the following "ln b" from above:

→ ln b = ln  [ = (0.5) ln (y² -1) ;

Rewrite:  →    ln(a * b) = ln(a) + ln(b) = ln 27 + ln (x³) ;

      = (3 ln 3 +  3 ln x) + [ln [ =

      = (3 ln 3 +  3 ln x) +  0.5 (ln [) ;

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So, this is the value for the "ln" of the "numerator" of the given problem:

" (3 ln 3 +  3 ln x) +  0.5 (ln [) " .

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Now, let's find the value for the "ln" of the "denominator" of the given problem:

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  →  " ln y(x -1)²  " ;

  =  ln * ;

→ Use the product rule:  ln(a * b) = ln a + ln b ;  a = "y" ;   b = " (x-1)² " .

→   ln (*)  = (ln y) * (ln (x-1)²) ;

Note:  ln (x-1)² = 2 ln (x-1) ;

→ (ln y) * [ ln (x-1)² ] =  (ln y ) * 2 ln (x-1) = 2 (ln y) [ ln (x-1) ] .

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So, going back to our original given problem:

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→   ;

→  Use the "quotient rule" property of the "natural logarithm (ln)" ;

          ln (a/b)  = ln a − ln b  ;

Note:  " ln a "  = "  (3 ln 3 +  3 ln x) +  0.5 (ln [)

           "ln b " =  " (2 (ln y) [ ln (x-1) ]

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   (ln a) - (ln b) =

          ((3 ln 3 +  3 ln x) +  0.5 (ln (y^{2}-1) −

    { (2 (ln y) [ ln (x-1) ]  ;

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answer: -2+√x,-2-√x

i hope this , i don't know if you're looking for a pair or not.