Solve log_(1/3)[log_(1/4) (x^2-3x)]<0 - ScanMath

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Question

Solve the inequality

x \in (\frac{- 10 + 3}{2}, 0) \cup (3, \frac{10 + 3}{2})

Evaluate

lo g_{\frac{1}{3}} (lo g_{\frac{1}{4}} (x^{2} - 3 x)) < 0

Find the domain

More Steps

lo g_{\frac{1}{3}} (lo g_{\frac{1}{4}} (x^{2} - 3 x)) < 0, x \in (\frac{- 13 + 3}{2}, 0) \cup (3, \frac{13 + 3}{2})

For 0< \frac{1}{3} <1 the expression lo g_{\frac{1}{3}} (lo g_{\frac{1}{4}} (x^{2} - 3 x)) < 0 is equivalent to lo g_{\frac{1}{4}} (x^{2} - 3 x) > (\frac{1}{3})^{0}

lo g_{\frac{1}{4}} (x^{2} - 3 x) > (\frac{1}{3})^{0}

Evaluate the power

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

For 0< \frac{1}{4} <1 the expression lo g_{\frac{1}{4}} (x^{2} - 3 x) > 1 is equivalent to x^{2} - 3 x < (\frac{1}{4})^{1}

x^{2} - 3 x < (\frac{1}{4})^{1}

Evaluate the power

x^{2} - 3 x < \frac{1}{4}

Add the same value to both sides

x^{2} - 3 x + \frac{9}{4} < \frac{1}{4} + \frac{9}{4}

Evaluate

x^{2} - 3 x + \frac{9}{4} < \frac{5}{2}

Evaluate

(x - \frac{3}{2})^{2} < \frac{5}{2}

Take the 2 -th root on both sides of the inequality

(x - \frac{3}{2})^{2} < \frac{5}{2}

Calculate

x - \frac{3}{2} < \frac{10}{2}

Separate the inequality into 2 possible cases

{x - \frac{3}{2} < \frac{10}{2} x - \frac{3}{2} > - \frac{10}{2}

Calculate

More Steps

{x < \frac{10 + 3}{2} x - \frac{3}{2} > - \frac{10}{2}

Calculate

More Steps

{x < \frac{10 + 3}{2} x > \frac{- 10 + 3}{2}

Find the intersection

\frac{- 10 + 3}{2} < x < \frac{10 + 3}{2}

Check if the solution is in the defined range

\frac{- 10 + 3}{2} < x < \frac{10 + 3}{2}, x \in (\frac{- 13 + 3}{2}, 0) \cup (3, \frac{13 + 3}{2})

Solution

x \in (\frac{- 10 + 3}{2}, 0) \cup (3, \frac{10 + 3}{2})

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