17 3^72cdot ((1)/(3))^xcdot ((1)/(3))^sqrt (x)gt 1

To solve the inequality $3^{72}\cdot (\frac {1}{3})^{x}\cdot (\frac {1}{3})^{\sqrt {x}}\gt 1$, we can simplify the expression and solve for $x$.<br /><br />First, let's rewrite the expression using the properties of exponents:<br /><br />$3^{72}\cdot (\frac {1}{3})^{x}\cdot (\frac {1}{3})^{\sqrt {x}} = 3^{72}\cdot 3^{-x}\cdot 3^{-\sqrt{x}}$<br /><br />Now, we can combine the exponents:<br /><br />$3^{72}\cdot 3^{-x}\cdot 3^{-\sqrt{x}} = 3^{72-x-\sqrt{x}}$<br /><br />So, the inequality becomes:<br /><br />$3^{72-x-\sqrt{x}} > 1$<br /><br />Since $3^{72-x-\sqrt{x}} > 1$, we know that the exponent must be greater than 0:<br /><br />$72-x-\sqrt{x} > 0$<br /><br />Now, let's solve this inequality for $x$:<br /><br />$72-x-\sqrt{x} > 0$<br /><br />Rearranging the terms, we get:<br /><br />$x + \sqrt{x} < 72$<br /><br />Let $y = \sqrt{x}$. Then, the inequality becomes:<br /><br />$y^2 + y < 72$<br /><br />Rearranging the terms, we get:<br /><br />$y^2 + y - 72 < 0$<br /><br />This is a quadratic inequality. We can solve it by factoring:<br /><br />$(y - 8)(y + 9) < 0$<br /><br />The solutions to this inequality are:<br /><br />$-9 < y < 8$<br /><br />Since $y = \sqrt{x}$, we can substitute back to get:<br /><br />$-9 < \sqrt{x} < 8$<br /><br />Squaring both sides, we get:<br /><br />$81 < x < 64$<br /><br />Therefore, the solution to the inequality $3^{72}\cdot (\frac {1}{3})^{x}\cdot (\frac {1}{3})^{\sqrt {x}}\gt 1$ is $81 < x < 64$.