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Question on exponent laws

  1. Jul 7, 2011 #1
    1. The problem statement, all variables and given/known data

    2^3 / 2^0 -2^-1

    2. Relevant equations
    solving using exponent laws



    3. The attempt at a solution

    i move the negetive one over and make it 2^3 /2^0 = 1/2^1
    I get 8 on the left side, and 2 on the right, bring the 6 from the right to the left and get 2. The answer is 16, i dont know how to get it. Anyone?
     
  2. jcsd
  3. Jul 7, 2011 #2
    2^3 / 2^0 -2^-1

    Is that:

    [tex]\frac{2^3}{2^{0}-2^{-1}}[/tex] ?
     
  4. Jul 7, 2011 #3

    fluidistic

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    Yes because the answer is 16 :)
     
  5. Jul 7, 2011 #4
    yes it isss
     
  6. Jul 7, 2011 #5

    HallsofIvy

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    [tex]\frac{2^3}{2^0}- 2^{-1}= \frac{8}{1}- \frac{1}{2}= 8- \frac{1}{2}= \frac{15}{2}[/tex]

    [tex]\frac{2^3}{2^0- 2^{-1}}= \frac{8}{1-\frac{1}{2}}= \frac{8}{\frac{1}{2}}= 16[/tex]
     
  7. Jul 7, 2011 #6
    How do you solve this problem?

    3^g+3 - 3^g+2 = 1458

    Ive been stuck on it for a whole hour
     
  8. Jul 7, 2011 #7

    Ray Vickson

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    There is no solution. You write 3^g+3 - 3^g+2, which means (3^g) + 3 - (3^g) + 2, and this is equal to 5. If you actually mean something different, USE BRACKETS. For example, maybe you mean 3^(g+3) - 3^(g+2) = 1458. Of course, that would have a solution, which we could get quite easily using the fact that 3^(g+3) = 3*3^(g+2).

    RGV
     
  9. Jul 7, 2011 #8
    Theres no brackets, and there is a solution , its 4.

    Heres another one.

    -500 = 5^y+1 -5^y+2

    looks like the same type of problem, answer to this ones 3
     
  10. Jul 7, 2011 #9

    Ray Vickson

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    You seem to be missing a VERY IMPORTANT point: you *must* use brackets because otherwise what you are writing is in violation of all standard math writing rules. People will automatically assume you mean what you write, but apparently that is not true!

    RGV
     
  11. Jul 7, 2011 #10

    SammyS

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    If you mean that y+1 and y+2 are exponents, then you should (must, actually) use parentheses or other grouping symbols to indicate where the exponent ends and the rest of the expression picks up again.

    I assume you meant:

    -500 = 5^(y+1) - 5^(y+2) .

    It works even better to use the X2 button above the advanced version of the reply box. Then you don't need the parentheses. This gives:

    -500 = 5y+1 - 5y+2

    To solve this, notice that 5y+2 = (5)(5y+1)

    Factor (5y+1) out of 5y+1 - (5)(5y+1).

    The answer is y = 2.
     
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