What is log_{6}(x^2)>= log_{6}(9) ?

The solution to log_{6}(x^2)>= log_{6}(9) is x = 3

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log_{6}(x^2)>= log_{6}(9)

lo g_{6} (x^{2}) \geq lo g_{6} (9)

x \leq - 3 or x \geq 3

Interval Notation

(- \infty, - 3] \cup [3, \infty)

Solution steps

lo g_{6} (x^{2}) \geq lo g_{6} (9)

Apply log rules

x^{2} \geq 9

x^{2} \geq 9 : x \leq - 3 or x \geq 3

x \leq - 3 or x \geq 3

lo g_{\frac{1}{2}} (lo g_{2} (\frac{x}{( 1 + x )})) > 0

lo g_{x} (\frac{2 x - 3}{4}) > 2

lo g_{4} (- x) < 0

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

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

What is log_{6}(x^2)>= log_{6}(9) ?
The solution to log_{6}(x^2)>= log_{6}(9) is x<=-3\lor x>= 3