Question

The solution set of $(\frac{2}{3})^{x^{2}-5x+9} > (\frac{27}{8})^{-1}$ is

Answer 1

(2,3)

Answer 2

The given inequality is: $(\frac{2}{3})^{x^{2}-5x+9} > (\frac{27}{8})^{-1}$

First, simplify the right-hand side of the inequality: $(\frac{27}{8})^{-1} = (\frac{3^3}{2^3})^{-1} = ((\frac{3}{2})^3)^{-1}$ Using the property $(a^m)^n = a^{mn}$ and $a^{-1} = \frac{1}{a}$: $((\frac{3}{2})^3)^{-1} = (\frac{3}{2})^{-3}$ To match the base on the left-hand side, we can write $\frac{3}{2}$ as $(\frac{2}{3})^{-1}$: $(\frac{3}{2})^{-3} = ((\frac{2}{3})^{-1})^{-3} = (\frac{2}{3})^{(-1) \times (-3)} = (\frac{2}{3})^3$

Now, substitute this back into the original inequality: $(\frac{2}{3})^{x^{2}-5x+9} > (\frac{2}{3})^3$

The base of the exponential function is $\frac{2}{3}$. Since $0 < \frac{2}{3} < 1$, the exponential function $f(t) = (\frac{2}{3})^t$ is a strictly decreasing function. Therefore, when comparing the exponents, the inequality sign must be reversed: $x^{2}-5x+9 < 3$

Now, solve this quadratic inequality: $x^{2}-5x+9 - 3 < 0$ $x^{2}-5x+6 < 0$

To find the values of $x$ for which this inequality holds, we first find the roots of the corresponding quadratic equation $x^{2}-5x+6 = 0$. We can factor the quadratic expression: $(x-2)(x-3) = 0$ The roots are $x=2$ and $x=3$.

Since the coefficient of $x^2$ is positive (which is 1), the parabola $y = x^{2}-5x+6$ opens upwards. For $x^{2}-5x+6 < 0$, the parabola must be below the x-axis. This occurs between its roots. Therefore, the solution for the inequality $x^{2}-5x+6 < 0$ is $2 < x < 3$.

The solution set is the open interval $(2, 3)$.