SOLUTION: I am having trouble with this could someone explain and show me how to work this problem.
Preform the operations. Write the ansers without negative exponents. Assume that all va
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Preform the operations. Write the ansers without negative exponents. Assume that all va
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Question 45456This question is from textbook intermediate algebra
: I am having trouble with this could someone explain and show me how to work this problem.
Preform the operations. Write the ansers without negative exponents. Assume that all varibles represent postive numbers.
(27x^-3)^-1/3 This question is from textbook intermediate algebra
You can put this solution on YOUR website! (27x^-3)^-1/3
The "negative" in the exponent means to invert.
The "1/3" in the exponent means to take the cube root.
=(27)^(-1/3)*(x^-3)^(-1/3)
= (3)^-1 *(x^1)
=(1/3) x
Cheers,
Stan H.
: I am having trouble with this could someone explain
and show me how to work this problem. Perform the
operations. Write the ansers without negative
exponents. Assume that all varibles represent
postive numbers.
(27x^-3)^-1/3
(27x-3)-1/3
Give the 27 a 1 exponent
(271x-3)-1/3
Multiply the outer exponent -1/3 by each of the
inner exponents:
271·-1/3x-3·-1/3
27-1/3x1
To get rid of the negative exponent, put the
whole thing over 1,
27-1/3x1
---------
1
then move the base and exponent from numerator
to denominator, and change the sign of
the exponent to positive:
x1
---------
1·271/3
Erase the 1 exponent in the numerator and the
1 multiplier in the denominator:
x
-------
271/3
Now the 1/3 power is the same as the cube
root, so now we have
x
-------
³Ö27
The cube root of 27 is 3, so the final
answer is
x
---
3
Edwin
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