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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`