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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%.

\((x-1)(x^2-7x+9)=0\)
  

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

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

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

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

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

Algebra, MET1 2015 VCAA 7b

Solve  `3e^t = 5 + 8e^(−t)`  for `t`.  (3 marks)

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`log_e(8/3)`

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`3e^t – 5 – 8e^(−t) = 0`

♦ Part (b) mean mark 44%.
MARKER’S COMMENT: Many students could not create a quadratic, and others who did made mistakes by trying to solve using the quadratic formula rather than simple factorising.

`text(Multiply both sides by)\ e^t:`

`3e^(2t) – 5e^t – 8 = 0`

`text(Let)\ \ y = e^t`

`log_2((6 – x)/(4 – x))` `= 2`
`3y^2 – 5y – 8` `= 0`
`(3y – 8)(y + 1)` `= 0`
`y` `=8/3\ quadquadquadquadquad text(or)\ \ \ \ ` `y` `=-1\ \ text{(No solution)}`
`e^t` `= 8/3`  
`:. t` `= log_e(8/3)\ \ \ `    

Algebra, MET1 SM-Bank 10

Solve the equation  `log_e x-3/log_ex=2`  for `x`.   (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 substitute `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`
COMMENT: Must know how to find `x` from the equations  `log_ex=-1`  and  `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`

Algebra, MET1 SM-Bank 9

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`

Algebra, MET1 2011 VCAA 2b

Solve the equation  `4^x - 15 × 2^x = 16`  for `x.`  (3 marks)

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`x = 4`

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`4^x – 15 xx 2^x – 16` `= 0`
`2^(2x) – 15 xx 2^x – 16` `= 0`

 

♦♦ Mean mark 33%.
MARKER’S COMMENT: “Poorly answered”. Many students incorrectly stated that if  `y=2^x`, then `2y=4^x`.

`text(Let)\ \ y = 2^x`

`y^2 – 15y – 16` `= 0`
`(y – 16) (y + 1)` `= 0`
`y` `= 16` `\ \ \ or\ \ \ ` `y` `= – 1`
`2^x` `= 16`   `2^x` `= – 1`
`:. x` `= 4`   `text(No solution)`