SOLUTION: I am missing (or got a block for !) some very basic knowledge of exponents. Question: X^(-2/3) = (1/9), solve for x What I did: Raise the power of both L and R to -3/2 in or

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Question 1171982: I am missing (or got a block for !) some very basic knowledge of exponents.
Question: X^(-2/3) = (1/9), solve for x
What I did: Raise the power of both L and R to -3/2 in order to get rid of the exponent of the variable X. Then the equation is
X^(-2/3)^(-3/2) = (1/9)^(-3/2) and it solves to X = +/- 27. Book says I got the right answer.
But it says while going through initial step of X^(-2/3)^(-3/2) = (1/9)^(-3/2)
it wants me to add a +/- in front of the (1/9) as
X^(-2/3)^(-3/2) = +/-(1/9)^(-3/2)
Why do I have to insert the +/- sign at this stage. Is it an accepted practice that you add +/- whenever you raise the power a negative fraction?....since it is going to require it anyway further down the steps?
Please help
Thank you
Answer by math_helper(2461) (Show Source):
You can put this solution on YOUR website!
The reason is when you take a square root of both sides of an equation, you now need the +/- on one side or else information is lost.

Example

= +/- (*)
= +/-
or
If we only kept the positive root of both sides at (*), one answer would have been lost.

In your specific problem, raising to the -3/2 is a raise tox the -3 followed by the square root operation (or square root followed by raise to the -3), so that's where the square root is creeping in.
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Follow up: oh, I see what you mean. I do not think the +/- needs to be added at that particular step, as you technically have not taken the sqrt(LHS) yet. Once you write x by itself, then +/- is needed on RHS. Remember though, it is the sqrt (multiple of 1/2 in the exponent) that is causing this need, not the fact that the exponent is negative.