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

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

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

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

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

\(\text{By definition:}\)

\(x^5=m^n\)

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

\(\Rightarrow B\)

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`

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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`

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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.)}`