Logarithmic Domains

Algebra Level 4

f ( x ) = log 0.3 ( log 2 x ) \large f(x)=\sqrt {\log_{0.3} (\log_{2} x}) Find the domain of the function above.

( 2 , ) (2, \infty ) [ 2 , ) [2, \infty ) R \mathbb R ( 1 , 2 ] (1,2 ] [ 1 , 2 ] [1,2 ] ( 0 , ) (0, \infty )

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1 solution

Aquilino Madeira
Jul 23, 2015

D f = { x : log 0 , 3 ( log 2 x ) 0 x > 0 log 2 x > 0 } D f = { x : log 2 x ( 0 , 3 ) 0 x > 0 x > 2 0 } D f = { x : x 2 x > 0 x > 1 } D f = ] 1 , 2 ] \begin{array}{l} {D_f} = \left\{ {x \in \Re :{{\log }_{0,3}}\left( {{{\log }_2}x} \right) \ge 0\; \wedge \;x > 0\; \wedge \;{{\log }_2}x > 0} \right\}\\ {D_f} = \left\{ {x \in \Re :{{\log }_2}x \le {{(0,3)}^0}\; \wedge \;x > 0\; \wedge \;x > {2^0}} \right\}\\ {D_f} = \left\{ {x \in \Re :x \le 2\; \wedge \;x > 0\; \wedge \;x > 1} \right\}\\ {D_f} = \left] {1,2} \right] \end{array}

I don't understand why the square root has to be >= 0. There is something implicit disallowing complex numbers, I guess. Is this standard in finding the domain?

I would have thought the correct range is (1, inf) since the square root can handle any domain (assuming imaginary numbers are allowed).

Mike Williamson - 3 years, 7 months ago

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