Rewrite #sqrt((x^2*2^x)/((2x)^3*3^(2x)# as #a*x^b*c^x# for #x>0#?

1 Answer
Sep 12, 2017

#sqrt((x^2 * 2^x)/((2x)^3 * 3^(2x)))=color(red)(1/(2sqrt(2))) * x^(color(red)(-3/2)) * color(red)((sqrt(2)/3))^x#

Explanation:

#sqrt((x^2 * 2^x)/((2x)^3 * 3^(2x)))#

#color(white)("XXX")=color(blue)(sqrt(x^2)/(sqrt((2x)^3)) * color(green)(sqrt(2^x)/sqrt(3^(2x)))#

Taking these components one-at-a-time (and assuming #x > 0#)

```````````````````````````````````````````

#color(blue)(sqrt(x^2))=color(brown)x#

#color(blue)(1/sqrt((2x)^3))=1/((2x)^3)^(1/2)=1/((2x)^(3/2))=1/(2^(3/2) * x^(3/2))=1/(2sqrt(2)*x^(3/2))=color(brown)(1/(2sqrt(2))x^(-3/2))#

#rarr color(white)("XXX")color(blue)(sqrt(x^2)/(sqrt((2x)^3 * 3^(2x)))) = color(brown)x * color(brown)(1/(2sqrt(2))x^(-3/2)) = color(red)(1/(2sqrt(2)) * x^(-1/2)#

~~~~~~~~~~~~~~~~~~~~~~~~~

#color(green)(sqrt(2^x))=color(brown)(sqrt(2)^(x))#

#color(green)(1/sqrt(3^(2x)))=1/(3^x)=color(brown)((1/3)^x)#

#rarr color(white)("XXX")color(green)(sqrt(2^x)/sqrt(3^(2x)))=color(red)((sqrt(2)/3)^x)#