Question

Expand the following logarithm completely

Answer 1

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

{ (2 (ln y) [ ln (x-1) ] ;

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Explanation:

Given problem: Expand the following logarithm completely:

→
`ln [\frac{27x^(3)\sqrt{y^(2)-1 }  }{y(x-1)^(2) } }`] ;

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To begin: Simplify:

→
`[\frac{27x^(3)\sqrt{y^(2)-1 }  }{y(x-1)^(2) } }`] ;

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

"
`(y^(2) - 1)^{(1)/(2) }` " ;

→ Note the following property of exponents:

`√(a) = a^{(1)/(2) }` ;
`{ a\geq 0}.`

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Then rewrite the expression as:

`ln [\frac{27x^(3)(y^(2)-1)^{(1)/(2) } }{y(x-1)^(2) } ]`

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

ln (a/b) = ln a − ln b ;

So; we have:

`ln [\frac{27x^(3)(y^(2)-1)^{(1)/(2) } }{y(x-1)^(2) } ]`;

↔

`ln [\frac{27x^(3)(y^(2)-1)^{(1)/(2) } } [\frac{27x^(3)(y^(2)-1)^{(1)/(2) } }<strong> </strong>;</p><p> </p><p>=  ln  [tex][\frac{27x^(3)(y^(2)-1)^{(1)/(2) } }` -

ln
`[\frac{27x^(3)(y^(2)-1)^{(1)/(2) } }` ;

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ln a = ln
`[{27x^(3)(y^(2)-1)^{(1)/(2) } }` ;

ln b = ln
`(y^(2)-1)^{(1)/(2) } }` ;

Simplify: ln a − ln b ;

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First, simplify: ln a ;

ln a = ln
`{27x^(3)(y^(2)-1)^{(1)/(2) } }` ;

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

→ ln(a * b) = ln(a) + ln(b) ;

→ ln (
`{27x^(3)` *
`(y^(2)-1)^{(1)/(2) } }` ) ;

in which:

a =
`{27x^(3)` ;

b =
`(y^(2)-1)^{(1)/(2) } }` ;

→ ln (a * b) =

(ln a + ln b ) =

= [ ln (
`{27x^(3)`)] + [ ln [
`(y^(2)-1)^{(1)/(2) } }` = ? ;

First, simplify: "ln a " ;

→ ln a = ln
`{27x^(3)` = ? ;

→ 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 (
`{27x^(3)`)] + [ ln [
`(y^(2)-1)^{(1)/(2) } }` = ? ;

→ Rewrite: (3 ln 3 + 3 ln x) + [ln [
`(y^(2)-1)^{(1)/(2) } }` = ? ;

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

→ ln b = ln [
`(y^(2)-1)^{(1)/(2) } }` = (0.5) ln (y² -1) ;

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

= (3 ln 3 + 3 ln x) + [ln [
`(y^(2)-1)^{(1)/(2) } }` =

= (3 ln 3 + 3 ln x) + 0.5 (ln [
`(y^(2)-1)`) ;

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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 [
`(y^(2)-1)`) " .

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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
`y`*
`(x-1)^(2) } ]` ;

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

→ ln (
`y`*
`(x-1)^(2) } ]`) = (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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→
`ln [\frac{27x^(3)(y^(2)-1)^{(1)/(2) } }{y(x-1)^(2) } ]` ;

→ 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 [
`(y^(2)-1)`)

"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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