Answer: ((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:
→ [tex]ln [\frac{27x^{3}\sqrt{y^{2}-1 } }{y(x-1)^{2} } }[/tex]] ;
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To begin: Simplify:
→ [tex][\frac{27x^{3}\sqrt{y^{2}-1 } }{y(x-1)^{2} } }[/tex]] ;
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First, within the "numerator" , rewrite: " √(y² - 1) " ; as:
" [tex](y^{2} - 1)^{\frac{1}{2} }[/tex] " ;
→ Note the following property of exponents:
[tex]\sqrt{a} = a^{\frac{1}{2} }[/tex] ; [tex]{ a\geq 0}.[/tex]
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Then rewrite the expression as:
[tex]ln [\frac{27x^{3}(y^{2}-1)^{\frac{1}{2} } }{y(x-1)^{2} } ][/tex]
Note the "quotient rule" property of the "natural logarithm (ln)" ;
ln (a/b) = ln a − ln b ;
So; we have:
[tex]ln [\frac{27x^{3}(y^{2}-1)^{\frac{1}{2} } }{y(x-1)^{2} } ][/tex];
↔
[tex]ln [\frac{27x^{3}(y^{2}-1)^{\frac{1}{2} } } [\frac{27x^{3}(y^{2}-1)^{\frac{1}{2} } } ;
= ln [tex][\frac{27x^{3}(y^{2}-1)^{\frac{1}{2} } }[/tex] -
ln [tex][\frac{27x^{3}(y^{2}-1)^{\frac{1}{2} } }[/tex] ;
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ln a = ln [tex][{27x^{3}(y^{2}-1)^{\frac{1}{2} } }[/tex] ;
ln b = ln [tex](y^{2}-1)^{\frac{1}{2} } }[/tex] ;
Simplify: ln a − ln b ;
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First, simplify: ln a ;
ln a = ln [tex]{27x^{3}(y^{2}-1)^{\frac{1}{2} } }[/tex] ;
→ Note the "product rule" for the "natural logarithm" {" ln " }:
→ ln(a * b) = ln(a) + ln(b) ;
→ ln ( [tex]{27x^{3}[/tex] * [tex](y^{2}-1)^{\frac{1}{2} } }[/tex] ) ;
in which:
a = [tex]{27x^{3}[/tex] ;
b = [tex](y^{2}-1)^{\frac{1}{2} } }[/tex] ;
→ ln (a * b) =
(ln a + ln b ) =
= [ ln ([tex]{27x^{3}[/tex])] + [ ln [ [tex](y^{2}-1)^{\frac{1}{2} } }[/tex] = ? ;
First, simplify: "ln a " ;
→ ln a = ln [tex]{27x^{3}[/tex] = ? ;
→ 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 ([tex]{27x^{3}[/tex])] + [ ln [ [tex](y^{2}-1)^{\frac{1}{2} } }[/tex] = ? ;
→ Rewrite: (3 ln 3 + 3 ln x) + [ln [ [tex](y^{2}-1)^{\frac{1}{2} } }[/tex] = ? ;
Now, find simplify the following "ln b" from above:
→ ln b = ln [[tex](y^{2}-1)^{\frac{1}{2} } }[/tex] = (0.5) ln (y² -1) ;
Rewrite: → ln(a * b) = ln(a) + ln(b) = ln 27 + ln (x³) ;
= (3 ln 3 + 3 ln x) + [ln [ [tex](y^{2}-1)^{\frac{1}{2} } }[/tex] =
= (3 ln 3 + 3 ln x) + 0.5 (ln [[tex](y^{2}-1)[/tex]) ;
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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 [[tex](y^{2}-1)[/tex]) " .
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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 [tex]y[/tex]*[tex](x-1)^{2} } ][/tex] ;
→ Use the product rule: ln(a * b) = ln a + ln b ; a = "y" ; b = " (x-1)² " .
→ ln ([tex]y[/tex]*[tex](x-1)^{2} } ][/tex]) = (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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→ [tex]ln [\frac{27x^{3}(y^{2}-1)^{\frac{1}{2} } }{y(x-1)^{2} } ][/tex] ;
→ 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 [[tex](y^{2}-1)[/tex])
"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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