Mathematics
OpenStudy (anonymous):

solve. logx(27)=-3/2

OpenStudy (lgbasallote):

change to exponential form

OpenStudy (anonymous):

i don't understand how to do that.

OpenStudy (anonymous):

3^3=27

OpenStudy (lgbasallote):

$\large x^{-\frac{3}{2}} = 27$

OpenStudy (lgbasallote):

raise both sides by -1

OpenStudy (lgbasallote):

to cancel the negative exponent of x

OpenStudy (anonymous):

still there?

OpenStudy (anonymous):

yes...trying to work it out and get a grasp on it.

OpenStudy (anonymous):

i don't understand. i am getting x=1/9

OpenStudy (lgbasallote):

remember... $\LARGE x^{\frac{m}{n}} = \sqrt[n]{x^m}$

OpenStudy (anonymous):

still clueless...i'm just not getting it.

OpenStudy (lgbasallote):

did you get $\Large \frac{1}{27^{\frac{3}{2}}}$

OpenStudy (anonymous):

no

OpenStudy (lgbasallote):

$\large x^{-\frac{3}{2}} = 27$ $\large (x^{-\frac{3}{2}})^{-1} = 27^{-1}$ $\large x^{\frac{3}{2}} = \frac{1}{27}$ lol sorry i got confused...is that what you got so far?

OpenStudy (phi):

to solve $x^{-\frac{3}{2}} = 27$ use the rule $(x^a)^b= x^{ab}$ raise both sides to to -2/3 power (this makes the left side just x)

OpenStudy (phi):

to make sense of $27^{-2/3}$ remember: minus means flip : $$1/27^{2/3}$$ 1/3 means take the cube root 2 means square

OpenStudy (anonymous):

im'm totally confused now

OpenStudy (phi):

start with $x^{-\frac{3}{2}}= 27$ raise both sides to -2/3 $(x^{-\frac{3}{2}})^{-\frac{2}{3}}= 27^{-\frac{2}{3}}$

OpenStudy (phi):

can you simplify the left side? You multiply the exponents. you get $x^1 = 27^{-\frac{2}{3} }$

OpenStudy (anonymous):

ok. so to solve the problem would it be x=1/9

OpenStudy (phi):

For the right side, we can write it as $(27^{\frac{1}{3}})^{-2}$ Do you see how? $27^{\frac{1}{3}}$ means take the cube root of 27. 27= 3*3*3 can you find the cube root?

OpenStudy (anonymous):

3^3

OpenStudy (phi):

yes 27= 3^3 and $(3^3)^{\frac{1}{3}}$ is ??

OpenStudy (phi):

we find $27^{\frac{1}{3}} = 3$ so now we have $(27^{\frac{1}{3}})^{-2} = 3^{-2}$

OpenStudy (anonymous):

1/cube root of 3

OpenStudy (anonymous):

i am so much more confused now

OpenStudy (phi):

step by step we have $(27^{\frac{1}{3}})^{-2} = 3^{-2} = ??$

OpenStudy (anonymous):

1/9

OpenStudy (phi):

Yes. x= 1/9 If you are confused, go back and try to follow the rules we used to get to the answer.

OpenStudy (anonymous):

ok...i will. i have to remember this before my final. thanks so much for the step by step.

OpenStudy (phi):

The very first step was to know how to change logx(27)=-3/2 into $x^{-3/2} = 27$

OpenStudy (phi):

btw, it looks like you got the answer at the very beginning.

OpenStudy (anonymous):

thax