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L&E, 2ADV E1 2023 HSC 8 MC

What is the solution of the equation `log _a x^3=b`, where a and b are positive constants?

  1. `x=b^(a/3)`
  2. `x=a^(b/3)`
  3. `x=b^a/3`
  4. `x=a^b/3`
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`B`

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`log_a x^3` `=b`  
`3log_a x` `=b`  
`log_a x` `=b/3`  
`:.x` `=a^(b/3)`  

 
`=>B`

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 2 MC

If  `y = log_a (7x - b) + 3`, then `x` is equal to

  1. `1/7 a^(y - 3) + b`
  2. `(y - 3)/(log_a(7 - b))`
  3. `1/7 (a^(y - 3) + b)`
  4. `a^(y - 3) - b/7`
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`C`

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`y – 3` `= log_a (7x – b)`
`a^(y – 3)` `= 7x – b`
`a^(y – 3) + b` `= 7x`
`:. x` `= 1/7 (a^(y – 3) + b)`

`=>   C`

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`

L&E, 2ADV E1 2009 HSC 1f

Solve the equation  `lnx=2`. Give you answer correct to four decimal places.  (2 marks) 

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`7.3891`

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MARKER’S COMMENT: Students are reminded to write answers to more decimal places than required before rounding up.
`ln x` `=2`
`log_e x` `=2`
`x` `=e^2`
  `=7.38905…`
  `=7.3891\ \ text{(to 4 d.p.)}`