How do you solve #12x ^ { - \frac { 1} { 2} } = 6x ^ { \frac { 3} { 2} }#?

3 Answers
Gió
Feb 1, 2017

I got #x=sqrt(2)#

Explanation:

We can write it as:
#12/x^(1/2)=6x^(3/2)#
rearrange it:
#x^(3/2)x^(1/2)=12/6#
#x^(3/2+1/2)=2#
#x^(4/2)=2#
#x^2=2#
and
#x=+-sqrt(2)#
We can check our two solutions to see if they work:

1) #x=sqrt(2)=2^(1/2)#
we get:
#12(2^(1/2))^(-1/2)=6(2^(1/2))^(3/2)#
#12*2^(-1/4)=6*2^(3/4)#
#12/(2^(1/4))=6*2^(3/4)#
rearrange:
#12/6=2^(1/4)*2^(3/4)#
#2=2^(4/4)# YES

2) #x=-sqrt(2)=-2^(1/2)#
we get:
#12(-2^(1/2))^(-1/2)=6(-2^(1/2))^(3/2)#
#12*-2^(-1/4)=6*-2^(3/4)#
#12/(-2^(1/4))=6*-2^(3/4)#
rearrange:
#12/6=-2^(1/4)*-2^(3/4)#
#2=-2^(4/4)# NO

Ratnaker Mehta
Feb 1, 2017

#x=sqrt2#.

Explanation:

We will solve this eqn. in #RR#.

#12x^(-1/2)=6x^(3/2) rArr 12/x^(1/2)=6x^(3/2)rArr12=6x^(3/2)x^(1/2)#

#rArr 12=6x^(3/2+1/2)=6x^2 rArr x^2=12/6=2#

# x=+-sqrt2#.

As in #RR," neither "(-sqrt2)^(-1/2)" nor "(-sqrt2)^(3/2)# is defined,

#x=+sqrt2" is the Soln."#

Barney V.
Feb 1, 2017

#x=sqrt2#

Explanation:

#12x^(-1/2)=6x^(3/2)#

multiply both sides by#1/12#

#:.x^(-1/2=6/12x^(3/2)#

#:.6/12x^(3/2)=x^(-1/2)#

divide both sides by #x^(3/2)#

#:.6/12=x^(-1/2)/x^(3/2)#

#:.x^(-1/2)-:x^(2/3)=x^(-1/2) xx x^(-2/3)=x^(-1/2-3/2)=x^(-2)#

#:.6/12=1/x^2#

cross multiply

#:.6x^2=12#

#:.x^2=12/6#

#:.x^2=2#

#:.x=sqrt2#

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