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Functions, MET1 2024 VCAA 6

Solve  \(2 \log _3(x-4)+\log _3(x)=2\)  for \(x\).   (4 marks)

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\(\dfrac{7 + \sqrt{13}}{2}\)

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\(2\log_3(x-4)+\log_3(x)\) \(=2\)
\(\log_3x(x-4)^2\) \(=2\)
\(3^2\) \(=x(x-4)^2\)
\(x(x^2-8x+16)-9\) \(=0\)
\(x^3-8x^2+16x-9\) \(=0\)
\(\text{Test }x=1\)  
\(1^3-8(1)^2+16(1)-9\) \(=0\)
\(\therefore\ x-1\ \text{is a factor} \)  

  

\((x-1)(x^2-7x+9)=0\)
  

\(\text{Using quadratic formula to solve}\ x^2-7x+9=0\)

\(x\) \(=\dfrac{-(-7)\pm\sqrt{(-7)^2-4(1)(9)}}{2(1)}\)
  \(=\dfrac{7\pm \sqrt{49-36}}{2}\)
  \(=\dfrac{7\pm \sqrt{13}}{2}\)

\( x=1, \dfrac{7- \sqrt{13}}{2}, \dfrac{7 + \sqrt{13}}{2}\)

  
\(\text{For }\log_3(x-4)\ \text{to exist}\ x>4\)

\(\therefore\ \dfrac{7 + \sqrt{13}}{2}\ \text{ is the only possible solution.}\)

Algebra, MET1 2020 VCAA 4

Solve the equation  `2 log_2(x + 5) - log_2(x + 9) = 1`.  (3 marks)

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`x = text{−1}`

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`2 log_2(x + 5) – log_2(x + 9)` `= 1`
`log_2(x + 5)^2 – log_2(x + 9)` `= 1`
`log_2(((x + 5)^2)/(x + 9))` `= 1`
`((x + 5)^2)/(x + 9)` `= 2`
`x^2 + 10x + 25` `= 2x + 18`
`x^2 + 8x + 7` `= 0`
`(x + 7)(x + 1)` `= 0`

 
`:. x = −1\ \ \ \ (x != text{−7}\ \ text(as)\ \ x > text{−5})`

Algebra, MET2-NHT 2019 VCAA 14 MC

If  `2log_e(x) - log_e(x + 2) = log_e(y)`, then  `x`  is equal to

  1.  `(y + sqrt{y^2 + 8y})/(2)`
  2.  `(y ± sqrt{y^2 + 8y})/(2)`
  3.  `(y ± sqrt{y^2 - 8y})/(2)`
  4.  `(-1 ± sqrt{4y - 7})/(2)`
  5.  `(-1 + sqrt{4y - 7})/(2)`
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`A`

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`2log_e x – log_e(x + 2)` `= log_e y`
`log_e ({x^2}/{x + 2})`  `= log_e y`

 
`y = (x^2)/(x + 2)`
 
`text(S) text(olve for) \ x :`

`x = (y + sqrt(y^2 + 8y))/(2) \ \ text(only,) \ \ ((y – sqrt{y^2 + 8y})/(2) < 0) `
 
`=> \ A`

Functions, MET1 2009 VCAA 9

Solve the equation  `2 log_e (x) - log_e (x + 3) = log_e (1/2)`  for `x.`  (4 marks)

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`:. x = 3/2`

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`text(Simplify using log laws:)`

MARKER’S COMMENT:
Less than a quarter of students achieved full marks here, with many failing to eliminate `x=-1` as an answer.
`log_e (x^2) – log_e (x + 3)` `= log_e (1/2)`
`log_e (x^2/(x + 3))` `= log_e (1/2)`
`x^2/(x + 3)` `= 1/2`
`2x^2` `= x + 3`
`2x^2 – x – 3` `= 0`
`(2x – 3) (x + 1)` `= 0`

 

`:. x = 3/2,\ \ \ x > 0`

Functions, MET1 2012 VCAA 7

Solve the equation  `2 log_e (x + 2) - log_e(x) = log_e (2x + 1)`,  where  `x > 0`, for `x.`  (3 marks)

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`x = 4`

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`text(Simplify using log laws:)`

`log_e (x + 2)^2 – log_e (x)` `= log_e (2x + 1)`
`log_e ((x + 2)^2/x)` `= log_e (2x + 1)`
`(x + 2)^2/x` `= 2x + 1`
`x^2 + 4x + 4` `= 2x^2 + x`
`x^2-3x-4` `= 0`
`(x-4)(x+1)` `= 0`
`x` `= 4 or -1,\ \ text(but)\ x > 0`
`:. x` `= 4`

Algebra, MET2 2013 VCAA 18 MC

Let  `g(x) = log_2(x),\ \ x > 0`

Which one of the following equations is true for all positive real values of  `x?`

  1. `2g (8x) = g (x^2) + 8`
  2. `2g (8x) = g (x^2) + 6`
  3. `2g (8x) = (g (x) + 8)^2`
  4. `2g (8x) = g (2x) + 6`
  5. `2g (8x) = g (2x) + 64`
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`B`

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`text(Consider Option)\ B:`

♦♦ Mean mark 35%.
`text(LHS)` `= 2g(8x)`
  `= 2log_2(8x)`
  `= 2log_2(8) + 2log_2(x)`
  `=2log_2 (2^3)+ 2log_2(x)`
  `= 6 + log_2(x^2)`
  `= g(x^2) + 6`

 
`=>   B`