2/3^x²+5x>1
To solve the inequality \( \frac{2}{3^{x^2 + 5x}} > 1 \), we will first manipulate the inequality to get it in a more manageable form.
Start by rewriting the inequality: \[ \frac{2}{3^{x^2 + 5x}} > 1 \]
To eliminate the fraction, multiply both sides by \( 3^{x^2 + 5x} \) (noting that since \(3^{x^2 + 5x}\) is positive for all real \(x\), the direction of the inequality does not change): \[ 2 > 3^{x^2 + 5x} \]
Now, we can express this as: \[ 3^{x^2 + 5x} < 2 \]
To solve for \( x^2 + 5x \), take the logarithm (base 3) of both sides: \[ x^2 + 5x < \log_3(2) \]
Let's denote \( c = \log_3(2) \), making the inequality: \[ x^2 + 5x - c < 0 \]
Now, we will solve the quadratic inequality \( x^2 + 5x - c < 0 \). To do this, we need to find the roots of the quadratic equation: \[ x^2 + 5x - c = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ x = \frac{-5 \pm \sqrt{25 + 4c}}{2} \]
The roots are: \[ x_1 = \frac{-5 - \sqrt{25 + 4c}}{2}, \quad x_2 = \frac{-5 + \sqrt{25 + 4c}}{2} \]
The quadratic \( x^2 + 5x - c \) will be less than 0 between the two roots \( x_1 \) and \( x_2 \). Thus, the solution to the inequality \( x^2 + 5x - c < 0 \) is: \[ x_1 < x < x_2 \]
Finally, substituting back for \( c = \log_3(2) \): \[ \frac{-5 - \sqrt{25 + 4\log_3(2)}}{2} < x < \frac{-5 + \sqrt{25 + 4\log_3(2)}}{2} \]
This gives you the solution set for the original inequality. You can now evaluate the numerical values of \( x_1 \) and \( x_2 \) using a calculator if needed.
