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L&E, 2ADV E1 2024 MET1 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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\(2\log_3(x-4)+\log_3(x)\) \(=2\)
\(\log_3x(x-4)^2\) \(=2\)
\(x(x-4)^2\) \(=3^2\)
\(x(x^2-8x+16)-9\) \(=0\)
\(x^3-8x^2+16x-9\) \(=0\)

 
\(\text{Find a factor}\ \ \Rightarrow\ \ \text{Test}\ \ x=1:\)

\(1^3-8(1)^2+16(1)-9=0\)

\(\therefore\ x-1\ \text{is a factor} \)

♦♦ Mean mark 36%.

\((x-1)(x^2-7x+9)=0\)
  

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

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

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

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

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

L&E, 2ADV E1 EQ-Bank 3

Solve the equation  \(4^x-2^{x+2}=32\), showing all working.   (3 marks)

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\(x=3\)

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  \(2^{2 x}-2^{x+2}\) \(=2^5\)
  \(2^{2 x}-2^2 \times 2^x-32\) \(=0\)

 
\(\text{Let} \ \ 2^x=X\)

  \(X^2-4X-32\) \(=0\)
  \((X-8)(X+4)\) \(=0\)

\(X\) \(=8\) \(\quad\text{or}\quad\) \(X\) \(=-4\)
\(2^x\) \(=8\)   \(2^x\) \(=-4\ \ \text{(no solution)}\)
\(x\) \(=3\)      

 
\(\therefore x=3\)

L&E, 2ADV E1 SM-Bank 15

Solve the following equation for \(x\):

\(2^{2x}=3(2^{x+1})-8\).   (3 marks)

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\(x=1\ \ \text{or} \ \ 2\)

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\(2^{2x}\) \(=3(2^{x+1})-8\)  
\(2^{2x}\) \(=3(2 \times 2^{x})-8\)  
\(0\) \(=2^{2x}-6\cdot2^{x}+8\)  

 
\(\text{Let}\ \ X=2^{x}\)

\(X^2-6X+8\) \(=0\)  
\((X-4)(X-2)\) \(=0\)  
\(X\) \(=4\ \ \text{or}\ \ 2\)  

 
\(2^{x}=4\ \ \Rightarrow\ \ x=2\)

\(2^{x}=2\ \ \Rightarrow\ \ x=1\)

L&E, 2ADV E1 2020 MET1 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})`

L&E, 2ADV E1 2019 NHT 4

Solve  `log_3(t)-log_3(t^2-4) = -1`  for  `t`.   (3 marks)

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`4 `

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`log_3(t)-log_3(t^2-4)` `= -1`
`log_3 ({t}/{t^2-4})` `= -1`
`(t)/(t^2-4)` `= (1)/(3)`
`t^2-4` `= 3t`
`t^2-3t-4` `= 0`
`(t-4)(t+ 1)` `= 0`

 
`:. t=4 \ \ \ (t > 0, \ t!= –1)`

L&E, 2ADV E1 2019 HSC 15a

Solve  `e^(2 ln x) = x + 6`   (2 marks)

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`x = 3 or -2`

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♦ Mean mark 47%.

`e^(2 ln x)` `= x + 6`
`ln e^(2 ln x)` `= ln (x + 6)`
`2 ln x` `= ln (x + 6)`
`ln x^2` `= ln (x + 6)`
`x^2` `= x + 6`
`x^2 – x – 6` `= 0`
`(x – 3) (x + 2)` `= 0`

 
`:. x = 3 \ \ (x>0)`

L&E, 2ADV E1 SM-Bank 11 MC

Solve the equation  `e^(4x) - 5e^(2x) + 4 = 0`  for `x`

  1. `x= 1, 4`
  2. `x= – 4, – 1`
  3. `x= 0, log_e 2`
  4. `x= – log_e 2, 0, log_e 2 `
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`C`

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`e^(4x) – 5e^(2x) + 4 = 0`

`text(Let)\ \ X=e^(2x)`

`X^2-5X+4` `=0`
`(X-4)(X-1)` `=0`
`X` `=4 or 1`

 

`:.e^(2x)` `=4` `e^(2x)` `=1`
`2x` `=log_e 4` `x` `=0`
`x` `=(2log_e 2)/2`    
  `=log_e 2`    

`=> C`

L&E, 2ADV E1 2004 HSC 6a

Solve the following equation for `x`:

`e^(2x) + 3e^x − 10 = 0`.  (2 marks)

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`x = ln 2`

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`e^(2x) + 3e^x − 10` `= 0`
`:. (e^x)^2 + 3e^x − 10` `= 0`

 
`text(Let)\ \ X = e^x,`

`:. X^2 + 3X – 10` `= 0`
`(X + 5)(X − 2)` `= 0`
`:. X =2 or -5`

 

`text(If)\ \ X` `=2`
`e^x` `=2`
`ln e^x` `=ln 2`
`x` `=ln 2`
`text(If)\ \ X` `=-5`
`e^x` `=-5\ \ \ text{(no solution)}`

 
 `:. x=ln 2`

L&E, 2ADV E1 2007 HSC 6a

Solve the following equation for `x`:

`2e^(2x)-e^x = 0`.   (2 marks)

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`x = ln\ 1/2`

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`text(Solution 1)`

`2e^(2x)-e^x = 0`

`text(Let)\ \ X = e^x:`

`2X^2-X` `= 0`
`X (2X-1)` `= 0`

 
`X = 0 or 1/2`

 
`text(When)\ e^x = 0\  =>\ text(no solution)`

`text(When)\ e^x = 1/2`

`ln e^x` `= ln\ 1/2`
`:. x` `= ln\ 1/2`

 

`text(Solution 2)`

`2e^(2x)-e^x` `=0`
`2e^(2x)` `=e^x`
`ln 2e^(2x)` `=ln e^x`
`ln 2 +ln e^(2x)` `=x`
`ln 2 + 2x` `=x`
`x` `=-ln2`
  `=ln\ 1/2`

L&E, 2ADV E1 2008 HSC 7a

Solve  `log_e x-3/log_ex=2`   (3 marks)

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`x=e^3\ \ text(or)\ \ e^-1`

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IMPORTANT: Students should recognise this equation as a quadratic, and the best responses substituted `log_ex` with a variable such as `X`.
`log_e x-3/(log_ex)` `=2`
`(log_ex)^2-3` `=2log_e x`
`(log_ex)^2-2log_ex-3` `=0`
   
`text(Let)\  X=log_ex`  
`:.\ X^2-2X-3` `=0`
`(X-3)(X+1)` `=0`
MARKER’S COMMENT: Many responses incorrectly stated that there is no solution to `log_ex=-1` or could not find `x` given `log_ex=3`.
`X` `=3` `\ \ \ \ \ \ \ \ \ \ ` `X` `=-1`
`log_ex` `=3` `\ \ \ \ \ \ \ \ \ \ ` `log_ex` `=-1`
`x` `=e^3` `\ \ \ \ \ \ \ \ \ \ ` `x` `=e^-1`

 

`:.x=e^3\ \ text(or)\ \ e^-1`