What is log_{1/4}(((35-x^2))/x)>= (-1)/2 ?

The solution to log_{1/4}(((35-x^2))/x)>= (-1)/2 is -7<= x<-sqrt(35)\lor 5<= x<sqrt(35)

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log_{1/4}(((35-x^2))/x)>= (-1)/2

lo g_{\frac{1}{4}} (\frac{( 35 - x ^{2} )}{x}) \geq \frac{- 1}{2}

- 7 \leq x < - 35 or 5 \leq x < 35

Interval Notation

[- 7, - 35) \cup [5, 35)

Decimal Notation

- 7 \leq x < - 5.91607 \dots or 5 \leq x < 5.91607 \dots

Solution steps

lo g_{\frac{1}{4}} (\frac{35 - x ^{2}}{x}) \geq \frac{- 1}{2}

Simplify

lo g_{\frac{1}{4}} (\frac{35 - x ^{2}}{x}) \geq - \frac{1}{2}

Apply log rules

\frac{35 - x ^{2}}{x} \leq 2 and \frac{35 - x ^{2}}{x} > 0

\frac{35 - x ^{2}}{x} \leq 2 and \frac{35 - x ^{2}}{x} > 0 : - 7 \leq x < - 35 or 5 \leq x < 35

- 7 \leq x < - 35 or 5 \leq x < 35

lo g_{3} (x - 2) \leq 2

lo g_{0.1} (4 x^{2} - 1) > lo g_{0.1} (3 x)

lo g_{2} (6 x + 2) > 0

0. 8^{l o g_{125} (x)} \geq 0. 8^{l o g_{10} (125)}

ln (x - e) > 0

What is log_{1/4}(((35-x^2))/x)>= (-1)/2 ?
The solution to log_{1/4}(((35-x^2))/x)>= (-1)/2 is -7<= x<-sqrt(35)\lor 5<= x<sqrt(35)