smc-726-10-Log – Product/Quotient Rule

Graphs, MET2 2021 VCAA 2 MC

The graph of `y = log_e(x) + log_e(2x)`, where `x > 0`, is identical, over the same domain, to the graph of

`y = 2log_e(1/2x)`
`y = 2log_e(2x)`
`y = log_e(2x^2)`
`y = log_e(3x)`
`y = log_e(4x)`

Show Answers Only

`C`

Show Worked Solution

`text(Using log laws:)`

`log_e (x) + log_e (2x) = log_e (2x^2) \ \ text{for} \ \ x > 0`

`=> C`

Algebra, MET1 2020 VCAA 4

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

Show Answers Only

`x = text{−1}`

Show Worked Solution

`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, MET1-NHT 2018 VCAA 4

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

Show Answers Only

`4 `

Show Worked Solution

`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)`

Algebra, MET2-NHT 2019 VCAA 14 MC

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

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

Show Answers Only

`A`

Show Worked Solution

`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`

Algebra, MET2 SM-Bank 4 MC

Which expression is equivalent to `4 + log_2 x?`

`log_2 (4x)`
`log_2 (16 + x)`
`4 log_2 (2x)`
`log_2 (16x)`
`log_2 (2x)`

Show Answers Only

`D`

Show Worked Solution

`4 + log_2 x`	`= log_2 16 + log_2 x`
	`= log_2 16 x`

`=> D`

Functions, MET1 2007 VCAA 2a

Solve the equation `log_e(3x + 5) + log_e(2) = 2`, for `x`. (2 marks)

Show Answers Only

`x = (e^2 – 10)/6`

Show Worked Solution

`text(Simplify using log laws:)`

`log_e(6x + 10)`	`=2`
`6x +10`	`=e^2`
`:.x`	`= (e^2 – 10)/6`

Functions, MET1 2009 VCAA 9

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

Show Answers Only

`:. x = 3/2`

Show Worked Solution

`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)

Show Answers Only

`x = 4`

Show Worked Solution

`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, MET1 2013 VCAA 5a

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

Show Answers Only

`18/25`
`- 6`

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`

Algebra, MET1 2014 VCAA 6

Solve `log_e(x) - 3 = log_e(sqrtx)` for `x`, where `x > 0`. (2 marks)

Show Answers Only

`e^6`

Show Worked Solution

`text(Simplify using log laws:)`

`log_e(x) – log_e(sqrtx)`	`= 3`
`log_e(x/sqrtx)`	`= 3`
`x/sqrtx xx sqrtx/sqrtx`	`=e^3`
`sqrt x`	`= e^3`
`:. x`	`= e^6`

Algebra, MET1 2015 VCAA 7a

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

Show Answers Only

`10/3`

Show Worked Solution

`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`

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?`

`2g (8x) = g (x^2) + 8`
`2g (8x) = g (x^2) + 6`
`2g (8x) = (g (x) + 8)^2`
`2g (8x) = g (2x) + 6`
`2g (8x) = g (2x) + 64`

Show Answers Only

`B`

Show Worked Solution

`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`