What is (0.3)^{2x^2-3x-6}<0.00243 ?

The solution to (0.3)^{2x^2-3x-6} (sqrt(97)+3)/4

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(0.3)^{2x^2-3x-6}<0.00243

(0.3)^{2 x^{2} - 3 x - 6} < 0.00243

x < \frac{- 97 + 3}{4} or x > \frac{97 + 3}{4}

Interval Notation

(- \infty, \frac{- 97 + 3}{4}) \cup (\frac{97 + 3}{4}, \infty)

Decimal Notation

x < - 1.71221 \dots or x > 3.21221 \dots

Solution steps

(0.3)^{2 x^{2} - 3 x - 6} < 0.00243

Apply exponent rules

2 x^{2} - 3 x - 6 > 5

2 x^{2} - 3 x - 6 > 5 : x < \frac{- 97 + 3}{4} or x > \frac{97 + 3}{4}

x < \frac{- 97 + 3}{4} or x > \frac{97 + 3}{4}

4^{x} < 2^{x} + 1 + 3

5^{1 - x} < 5^{4 - 5 x}

x, (\frac{3}{2})^{1 + r \cdot l^{2}} > (\frac{2}{3})^{3 x - 1}

0. 3^{7} > 0. 3^{x^{2} + 6 x}

5^{3 - x^{2}} \leq 5^{2}

What is (0.3)^{2x^2-3x-6}<0.00243 ?
The solution to (0.3)^{2x^2-3x-6}<0.00243 is x<(-sqrt(97)+3)/4 \lor x>(sqrt(97)+3)/4