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This module provides practice problems designed to develop concepts related to fractional exponents.

We have come up with the following definitions.

  • x 0 1
  • x a = 1 x a size 12{ { {1} over {x rSup { size 8{a} } } } } {}
  • x a b = x a b size 12{ nroot { size 8{b} } {x rSup { size 8{a} } } } {}

Let’s get a bit of practice using these definitions.

Check all of your answers above on your calculator. If any of them did not come out right, figure out what went wrong, and fix it!

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Solve for x : x x size 12{ { {x rSup { size 8{ {3} wideslash {2} } } } over {x rSup { size 8{ {1} wideslash {2} } } } } } {} 17 1 2 17 1 2

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Simplify: x x size 12{ { {x} over { sqrt {x} } } } {}

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Simplify: x + x x + 1 x size 12{ { {x rSup { size 8{ {3} wideslash {2} } } + sqrt {x} } over {x rSup { size 8{ {5} wideslash {2} } } + { {1} over { sqrt {x} } } } } } {}

Multiply the top and bottom by x 1 2 .
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Now…remember inverse functions? You find them by switching the x and the y and then solving for y . Find the inverse of each of the following functions. To do this, in some cases, you will have to rewrite the things. For instance, in #9, you will start by writing y x 1 2 . Switch the x and the y , and you get x y 1 2 . Now what? Well, remember what that means: it means x y . Once you’ve done that, you can solve for y , right?

x 3

  • Find the inverse function.
  • Test it.
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x -2

  • Find the inverse function.
  • Test it.
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x 0

  • Find the inverse function.
  • Test it.
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Can you find a generalization about the inverse function of an exponent?

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Graph y 2 x by plotting points. Make sure to include both positive and negative x values.

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Graph y 2 2 x by doubling all the y-values in the graph of y 2 x .

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Graph y 2 x 1 by taking the graph y 2 x and “shifting” it to the left by one.

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Graph y 1 2 x by plotting points. Make sure to include both positive and negative x values.

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Questions & Answers

a perfect square v²+2v+_
Dearan Reply
kkk nice
Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
y=10×
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
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rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
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No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
is it 3×y ?
Joan Reply
J, combine like terms 7x-4y
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im not good at math so would this help me
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how did I we'll learn this
Noor Reply
f(x)= 2|x+5| find f(-6)
Prince Reply
f(n)= 2n + 1
Samantha Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
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preparation of nanomaterial
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
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Source:  OpenStax, Advanced algebra ii: activities and homework. OpenStax CNX. Sep 15, 2009 Download for free at http://cnx.org/content/col10686/1.5
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