What is (3^{x/2})^{x-1}>= (3/9)^{x-3} ?

The solution to (3^{x/2})^{x-1}>= (3/9)^{x-3} is x>= (3+a_{4})/(a_{4)+1}

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(3^{x/2})^{x-1}>= (3/9)^{x-3}

(3^{\frac{x}{2}})^{x - 1} \geq (\frac{3}{9})^{x - 3}

x \geq \frac{3 + a _{4}}{a _{4} + 1}

Solution steps

(3^{\frac{x}{2}})^{x - 1} \geq (\frac{3}{9})^{x - 3}

Apply exponent rules

a_{4} (x - 1) \geq - (x - 3)

a_{4} (x - 1) \geq - (x - 3) : x \geq \frac{3 + a _{4}}{a _{4} + 1}

x \geq \frac{3 + a _{4}}{a _{4} + 1}

3^{x} \geq 5

\frac{( ( 3 ^{x} - 5 ^{x} ) ( x - 2 ) )}{( x ^{2} + 5 x + 2 )} \geq 0

lo g_{x} (\frac{( 4 x + 5 )}{( 6 - 5 x )}) < - 1

lo g_{10} (x) + lo g_{10} (x + 3) < 1

lo g_{x} (4 x - 3) \geq 2

What is (3^{x/2})^{x-1}>= (3/9)^{x-3} ?
The solution to (3^{x/2})^{x-1}>= (3/9)^{x-3} is x>= (3+a_{4})/(a_{4)+1}