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L&E, 2ADV E1 SM-Bank 4

Solve the following equation for \(x\):

\(\log _3(x-4)-\log _3 x=\dfrac{4}{3} \log _3 8\)   (3 marks)

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\(x=-\dfrac{4}{15}\)

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\(\log _3(x-4)-\log _3 x\) \(=\dfrac{4}{3} \log _3 8\)
\(\log _3\left(\dfrac{x-4}{x}\right)\) \(=\log _3 8^{\frac{4}{3}}\)
\(\log _3\left(\dfrac{x-4}{x}\right)\) \(=\log _3 16\)
\(\dfrac{x-4}{x}\) \(=16\)
\(x-4\) \(=16x\)
\(15 x\) \(=-4\)
\(x\) \(=-\dfrac{4}{15}\)

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 2017 HSC 5 MC

It is given that  `ln a = ln b-ln c`, where  `a, b, c > 0.`

Which statement is true?

  1. `a = b-c`
  2. `a = b/c`
  3. `text(ln)\ a = b/c`
  4. `text(ln)\ a = (text(ln)\ b)/(text(ln)\ c)`
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`B`

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Mean mark 51%.
COMMENT: Use of log laws here proved difficult for many students.
`ln a` `= ln b-ln c`
`ln a` `= ln (b/c)`
`:. a` `= b/c`

 
`=>  B`

L&E, 2ADV E1 SM-Bank 9

Solve  `log_2(6-x)-log_2(4-x) = 2`  for `x`, where  `x < 4`.  (2 marks)

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`10/3`

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`text(Simplify using log laws:)`

`log_2((6-x)/(4-x))` `= 2`
`2^2` `= (6-x)/(4-x)`
`16-4x` `= 6-x`
`3x` `= 10`
`:. x` `= 10/3`

L&E, 2ADV E1 SM-Bank 7

Solve the equation  `2 log_3(5)-log_3 (2) + log_3 (x) = 2`  for  `x.`  (2 marks)

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`18/25`

Show Worked Solution
`log_3 (5)^2-log_3 (2) + log_3 (x)` `= 2`
`log_3 (25x)-log_3 (2)` `=2`
`log_3 ((25 x)/2)` `= 2`
`(25x)/2` `= 3^2`
`:. x` `= 18/25`