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Logarithm, SMB-022

Solve  `4^(x-1)=84`  for `x`, giving your answer correct to 1 decimal place.  (2 marks)

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`4.2`

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`4^(x-1)` `=84`  
`log_10 4^(x-1)` `=log_10 84`  
`(x-1)log_10 4` `=log_10 84`  
`x-1` `=(log_10 84)/(log_10 4)`  
`x` `=(log_10 84)/(log_10 4)+1`  
  `=4.196…`  
  `=4.2\ text{(to 1 d.p.)}`  

Logarithm, SMB-021

Solve  `3^a=28`  for `a`, giving your answer correct to 2 decimal places.  (2 marks)

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`3.03`

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`3^a` `=28`  
`log_10 3^a` `=log_10 28`  
`a xx log_10 3` `=log_10 28`  
`a` `=(log_10 28)/(log_10 3)`  
  `=3.033…`  
  `=3.03`  

Logarithm, SMB-020

Solve  `4^(x+1)=32`  for `x`.  (2 marks)

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`3/2`

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`4^(x+1)` `=32`  
`log_10 4^(x+1)` `=log_10 32`  
`(x+1) log_10 4` `=log_10 32`  
`x+1` `=(log_10 32)/(log_10 4)`  
`x` `=5/2-1`  
  `=3/2`  

Logarithm, SMB-018

Evaluate  `log_a 6`  given  `log_a 2=0.62`  and  `log_a 24=2.67`.  (2 marks)

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`1.43`

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`log_a 6` `=log_a (24/4)`  
  `=log_a 24-log_b 2^2`  
  `=log_a 24-2log_a 2`  
  `=2.67-2 xx 0.62`  
  `=1.43`  

Logarithms, SMB-010

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

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`x = text{−1}`

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`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 2008 HSC 7a

Solve  `log_2 x-3/log_2 x=2`   (3 marks)

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`x=8\ \ text(or)\ \ 1/2`

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IMPORTANT: Students should recognise this equation as a quadratic, and the best responses substituted `log_2 x` with a variable such as `X`.
`log_2 x-3/(log_2 x)` `=2`
`(log_2 x)^2-3` `=2log_2 x`
`(log_2 x)^2-2log_2 x-3` `=0`
   
`text(Let)\  X=log_2 x`  
`:.\ X^2-2X-3` `=0`
`(X-3)(X+1)` `=0`
MARKER’S COMMENT: Many responses incorrectly stated that there is no solution to `log_2 x=-1` or could not find `x` given `log_2 x=3`.
`X` `=3` `\ \ \ \ \ \ \ \ \ \ ` `X` `=-1`
`log_2 x` `=3` `\ \ \ \ \ \ \ \ \ \ ` `log_2 x` `=-1`
`x` `=2^3=8` `\ \ \ \ \ \ \ \ \ \ ` `x` `=2^{-1}=1/2`

 

`:.x=8\ \ text(or)\ \ 1/2`

Logarithms, SMB-006 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))`
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`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`

Logarithms, SMB-005

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)`

Logarithms, SMB-001 MC

It is given that  `log_10 a = log_10 b-log_10 c`, where  `a, b, c > 0.`

Which statement is true?

  1. `a = b-c`
  2. `a = b/c`
  3. `log_10 a = b/c`
  4. `log_10 a = (log_10 b)/(log_10 c)`
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`B`

Show Worked Solution
Mean mark 51%.
COMMENT: Use of log laws here proved difficult for many students.
`log_10 a` `= log_10 b-log_10 c`
`log_10 a` `= log_10 (b/c)`
`:. a` `= b/c`

 
`=>  B`

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`

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`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`