solve. logx(27)=-3/2

change to exponential form

i don't understand how to do that.

3^3=27

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

raise both sides by -1

to cancel the negative exponent of x

still there?

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

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

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

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

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

no

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

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)

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

im'm totally confused now

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

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

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

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?

3^3

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

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

1/cube root of 3

i am so much more confused now

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

1/9

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

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

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

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

thax

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