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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))`
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`

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 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`