Help: Simple negative exponents

CoKun

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Oct 4, 2013
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Please show me the steps on how to do this:
x=3/5
3x^-3 - 5x^-2 = 0

Thanks !
 
Re: Simple negative exponents

Step1: Insert the values
3(3/5)^-3 - 5(3/5)^-2 = 0
Step2: Get rid of the negative exponents by putting it in the denominator
3/(3/5)^3 - 5/(3/5)^2 = 0
Step3: Do the powers
3/(27/125) - 5/(9/25) = 0
Step4: Get rid of the complex fraction by flipping the denominator and multiply it to the numerator
3(125/27) - 5(25/9) = 0
Step5: Multiplication of the numerators
375/27 - 125/9 = 0
Step6: LCD is 27 so multiply 125/9 by 3/3
375/27 - 125(3)/9(3) = 0
375/27 - 375/27 = 0

Hope this helps ^ ^
 
Step1: Insert the values


3(3/5)^-3 - 5(3/5)^-2 = 0


Step2: Get rid of the negative exponents by putting it in the denominator < < <

3/(3/5)^3 - 5/(3/5)^2 = 0


1) Conan, if you had made up your own similar problem and had worked it out in detailed fashion
in your post instead, then you would have been helping without doing the student's entire home-
work problem.

2) You made things more complicated here by introducing complex fractions. Instead, I would
have highly recommended taken advantage of eliminating the negative portion of the exponents
by inverting the fractional bases:


3(3/5)^(-3) - 5(3/5)^(-2) = 0 \(\displaystyle \ \implies \)


3(5/3)^3 - 5(5/3)^2 = 0



(But then the student should be trying to complete the remaining steps after any hints or beginning steps were shown.)
 
To Denis and lookagain:

Thank you for pointing that out :)
I overdid it. Lesson learned ^ ^
 
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