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L&E, 2ADV E1 EQ-Bank 6

Solve the following equation for \(a\):

\(a^{\log_e 3}=9\)    (3 marks)

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\(a=e^2\)

Show Worked Solution
\(a^{\log_e 3}\) \(=9\)   
\(\log_e  a^{ \log_e 3}\) \(=\log_e  3^2\)
\(\log_e 3 \times \log_e a\) \(=2 \log_e  3\)
\(\log _e a\) \(=\dfrac{2 \log _e 3}{\log _e 3}\)
\(\log _e a\) \(=2\)
\(a\) \(=e^2\)

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

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

L&E, 2ADV E1 SM-Bank 3 MC

Given \(m\) and \(n\) are positive constants, which expression is equal to

\(\log _m x^5=n\)

  1. \(x=n^{\frac{m}{5}}\)
  2. \(x=m^{\frac{n}{5}}\)
  3. \(x=\dfrac{n^m}{5}\)
  4. \(x=\dfrac{m^n}{5}\)
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\(\Rightarrow B\)

Show Worked Solution

\(\log _m x^5=n\)

\(\text{By definition:}\)

\(x^5\) \(=m^n\)
\(x\) \(=\left(m^n\right)^{\frac{1}{5}}\)
  \(=m^{\frac{n}{5}}\)

 
\(\Rightarrow B\)

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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\(\dfrac{7 + \sqrt{13}}{2}\)

Show Worked Solution

\(2\log_3(x-4)+\log_3(x)\) \(=2\)
\(\log_3x(x-4)^2\) \(=2\)
\(x(x-4)^2\) \(=3^2\)
\(x(x^2-8x+16)-9\) \(=0\)
\(x^3-8x^2+16x-9\) \(=0\)

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

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

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

♦♦ 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 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}`

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})`

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 `

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)`

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)`
Show Answers Only

`B`

Show Worked Solution
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`

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`

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`

L&E, 2ADV E1 SM-Bank 6 MC

The expression

`log_c(a) + log_a(b) + log_b(c)`

is equal to

  1. `1/(log_c(a)) + 1/(log_a(b)) + 1/(log_b(c))`
  2. `1/(log_a(c)) + 1/(log_b(a)) + 1/(log_c(b))`
  3. `-1/(log_a(b))-1/(log_b(c))-1/(log_c(a))`
  4. `1/(log_a(a)) + 1/(log_b(b)) + 1/(log_c(c))`
Show Answers Only

`B`

Show Worked Solution

`text(Solution 1)`

`text(Using Change of Base:)`

`log_c(a) + log_a(b) + log_b(c)`

`=(log_a(a))/(log_a(c)) + (log_b(b))/(log_b(a)) + (log_c(c))/(log_c(b))`

`=1/(log_a(c)) + 1/(log_b(a)) + 1/(log_c(b))`

 
`=> B`

 

`text(Solution 2)`

`text(Let)\ \ x` `=log_c(a)`
`c^x` `=a`
`x log_a c` `=log_a a`
`x` `=1/log_a c`

 

`text(Apply similarly for the other terms.)`

`=> B`

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`
Show Answers Only

`C`

Show Worked Solution
`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 2016 HSC 14e

Write  `log 2 + log 4 + log 8 + … + log 512`  in the form  `a log b`  where `a` and `b` are integers greater than `1.`  (2 marks)

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`45 log 2`

Show Worked Solution

`log 2 + log 4 + log 8 + … + log 512`

`= log 2^1 + log 2^2 + log2^3 + … + log 2^9`

`= log 2 + 2 log 2 + 3 log 2 + … + 9 log 2`

`= 45 log 2`

♦ Mean mark 40%.
 
TIP: Note that `log 2 = log_10 2`

L&E, 2ADV E1 2005 HSC 5a

Use the change of base formula to evaluate  `log_3 7`, correct to two decimal places.  (1 mark)

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`1.77\ \ text{(to 2 d.p.)}`

Show Worked Solution
`log_3 7` `= (log_10 7)/(log_10 3)`
  `= 1.771…`
  `= 1.77\ \ text{(to 2 d.p.)}`

L&E, 2ADV E1 2014 HSC 3 MC

What is the solution to the equation  `log_2(x-1) = 8`? 

  1. `4`
  2. `17`
  3. `65`
  4. `257`
Show Answers Only

`D`

Show Worked Solution
`log_2 (x-1)` `= 8`
`x-1` `= 2^8`
`x` `= 257`

 
`=>  D`

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`

Show Worked Solution
 
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`

Show Worked Solutions
MARKER’S COMMENT: Write answers to enough decimal places before rounding up.
`ln x` `=2`
`log_e x` `=2`
`x` `=e^2`
  `=7.38905…`
  `=7.3891\ \ text{(to 4 d.p.)}`

L&E, 2ADV E1 2013 HSC 9 MC

What is the solution of   `5^x=4`?

  1. `x=(log_2 4)/5`
  2. `x=4/(log_2 5)`
  3. `x=(log_2 4)/(log_2 5)`
  4. `x=log_2(4/5)`
Show Answer Only

 `C`

Show Worked Solutions
`5^x` `=4`
`log_2 5^x` `=log_2 4`
`x  log_2 5` `=log_2 4`
`:.x` `=(log_2 4)/(log_2 5)`

 
`=>C`