[sqrt {3}^(x+2)]/[3^(3x)]=3^{-x^2}

[jonboy]

So we have:\(\displaystyle \L \;\frac{\sqrt{3}^{x\,+\,2}}{27^x}\,=\,3^{-\,x^2}\)

Simplify 27 to 3³ and square both sides of equation:\(\displaystyle \L \;\frac{3^{(x\,+\,2)}}{3^{2(3x)}}\,=\,3^{2(-\,x^2)}\)

Now:\(\displaystyle \;x\,+\,2\,-\,6x\,=\,-\,2x^2\)


Simplify:\(\displaystyle \L \;2x^2\,-\,5x\,+\,2\,=\,0\)

So now complete the square?

 

[galactus]

You could just factor. Complete the square if you like.

 

[jonboy]

You could just factor. Complete the square if you like.

Yeah factoring sounds best. I got \(\displaystyle (\,-\,2x\,+\,1)\,(\,-\,x\,+\,2)\,\Rightarrow\,x\,=\,2\,,\,\frac{1}{2}\)

 

[stapel]

Simplify 27 to 3³ and square both sides of equation....

Another method: Note that the square root is the one-half power, and convert:

. . . . .3^(1/2)(x+2) / 3^3x = 3^{-x^2}

. . . . .3^{x/2 + 1 - 3x} = 3^{-x^2}

. . . . .3^{1 - (5/2)x} = 3^{-x^2}

Then:

. . . . .1 - (5/2)x = -x²

. . . . .x² - (5/2)x + 1 = 0

Note: The Quadratic Formula, derived from the completing-the-square process, is another method of solving the resulting quadratic equation.

Eliz.