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Algebra, MET2 2020 VCAA 10 MC

Given that  `log_(2)(n+1)=x`, the values of  `n`  for which  `x`  is a positive integer are

  1. `n=2^(k),k inZ^(+)`
  2. `n=2^(k)-1,k inZ^(+)`
  3. `n=2^(k-1),k inZ^(+)`
  4. `n=2k-1,k inZ^(+)`
  5. `n=2k,k inZ^(+)`
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`B`

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`text(If)\ \ x in ZZ^+ ,`

`log_2(n+1)` `in ZZ^+`  
`n+1` `=2^k\ \ text(for)\ \ k in ZZ^+`  
`n` `=2^k -1`  

 
`=>B`

Algebra, MET2 2007 VCAA 3 MC

If  `y = log_a (7x - b) + 3`, then `x` is equal to

  1. `1/7 a^(y - 3) + b`
  2. `1/7 (a^y - 3) + b`
  3. `1/7 (a^(y - 3) + b)`
  4. `a^(y - 3) - b/7`
  5. `(y - 3)/(log_a(7 - b))`
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`C`

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`y – 3` `= log_a (7x – b)`
`a^(y – 3)` `= 7x – b`
`a^(y – 3) + b` `= 7x`
`:. x` `= 1/7 (a^(y – 3) + b)`

 
`=>   C`

Algebra, MET1 SM-Bank 10

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

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`x=e^3\ \ text(or)\ \ e^-1`

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IMPORTANT: Students should recognise this equation as a quadratic, and substitute `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`
COMMENT: Must know how to find `x` from the equations  `log_ex=-1`  and  `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`