Question

please answer$\frac{\sqrt[3]{9^{x-5} } }{27}\leq \frac{\sqrt{1} }{\sqrt{3x^{x+2} } }$

Answer 1

I don't like inequalities, but you said please.

This is a toughie for middle school.  I'm going to guess the denominator on the right is really $\sqrt{3^{x+2}}$

We note we're dealing with all positive numerators and denominators, because the radical sign on the square root indicates the principal value of the square root which is always positive (or imaginary, in which case inequalities don't apply).  The cube root is of a positive number so will be positive.

Cross multiplying doesn't change the sense of the inequality because the factors are positive.  Of course the square root of 1 is 1.

$\sqrt[3]{9^{x-5}} \sqrt{3^{x+2}} \le 27$

Rewriting the radicals as fractional exponents,

${(9^{(x-5)}})^{\frac 1 3} {(3^{(x+2)})}^{\frac 1 2}\le 3^3$

Let's raise everything to the sixth power to get rid of those fractional exponents.

${(9^{(x-5)})}^{2} {(3^{(x+2)})}^3 \le 3^{18}$

${(3^2)^{2x-10}}{3^{3x+6}} \le 3^{18}$

${3^{4x-20}} {3^{3x+6}} \le 3^{18}$

Everything's a power of 3, so we take the log base 3.  Log is monotonic increasing so the inequality is preserved.

$4x-20 + 3x+6 \le 18$

$7x -14 \le 18$

$7x \le 32$

$x \le \dfrac{32}{7}$