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Need ALGEBRA 2 HELP with square roots

  1. May 3, 2008 #1
    1. The problem statement, all variables and given/known data
    simplify m^(9*√5)/m^(√5)

    3. The attempt at a solution
    would that equal m^9 or m^(8*√5)
     
  2. jcsd
  3. May 3, 2008 #2

    symbolipoint

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    You are confused by your own notation. If you mean the numerator is m^(9*(5)^0.5) and the denominator is m^((5)^(.5)), then the simplification is m^(8*(5)^(.5)). This is your second alternative which you are deciding.
     
  4. May 3, 2008 #3
    ahhh hold on let me write it again

    m ^ (9 X √5) DIVIDED by m ^ (√5) this problem involves properties of exponents and its a multiple choice problem so i know the answer can not be this m^(8*(5)^(.5)).

    Also i need help solving this:
    1. 5Log 5 65
    2. Log (1/3)x = -1
    3. 5^((log 5 2x - log 5 (x-3)) = ln e x+4
    4. 63n = 435n-4
    5. 4 + 3e5x = 27
     
  5. May 3, 2008 #4
    please anyone?
     
  6. May 3, 2008 #5
    x^y/x^z = x^(y-z) Dividing two exponents is subtracting their exponents. As for the logarithm problems, you might want to look up the properties of logs and exponents and play around with them.
     
  7. May 3, 2008 #6
    you didnt even address how to do things with square roots in them and i figured out how to do 4 and 5 :)
     
  8. May 3, 2008 #7
    I think it's fine to leave the square roots alone, unless you want to give a non-exact value. The second answer in your original post looks right.
     
  9. May 3, 2008 #8
  10. May 3, 2008 #9
    right now im stuck on this problem (1/6)^x = 36^(x+3) if i solve it by making the bases equal to 36 by multiplying 1/6 by -2 i get -2x but then for x i get -3 which obviously would not work, but if i use -2 as x the equation works BUT how do i get -2 from that equation????
     
  11. May 3, 2008 #10
    and thanks for your help with the first one :) really appreciate it
     
  12. May 3, 2008 #11
    I don't understand what you mean by making the base equal to 36. What I would do is use logarithms. If a = b, then ln(a) = ln(b). Then you can use the log identities to simplify and solve for x.

    No problem.
     
  13. May 3, 2008 #12
    my alg2 teacher said inorder to solve (1/6)^x = 36^(x+3) you need to make the BASES equal to each other i can easily make 1/6 to 36 by raising it to -2 but then i have to multiply x by -2. How do the log identities work?
     
  14. May 3, 2008 #13

    symbolipoint

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    Use the base of 6. On the leftside, you have 6^(-x). On the rightside you have (6^2)^(x+3)=6^(2(x+3))

    You then simply have 6^(-x) = 6^(2(x+3))

    *Understand the reasoning that lead to that.
    *Finish the solution process.
     
  15. May 3, 2008 #14
    yes symboli but then i get x = -1 and i know that x has to be -2, before when i made both bases equal to 36 i got -3 and that makes 36^0. so how do i get the equation to get me -2 for x?
     
  16. May 3, 2008 #15
    no i said
    if i raise 1/6 to -2 then it equals 36
     
  17. May 3, 2008 #16
    but doing that and what symboli says only gets me -3 or -1 not -2 :(?!?
     
  18. May 3, 2008 #17
    If you use what symbolipoint wrote, you get -x = 2(x+3). Then you just solve for x.

    Or if you use base 36, then 1/6 = 36^-.5, not -2.

    Edit: x = -2 if you use base 6 (doing what symboli says).
     
  19. May 3, 2008 #18
    lol i just did it and got it this is why i should not use a pen with math :P
     
  20. May 3, 2008 #19
    now if only i knew how to do this 6^(3n) = 43^(5n-4)
     
  21. May 4, 2008 #20

    HallsofIvy

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    You don't need log identities, just the fact that the exponential functions (ax for any positive a) are "one-to-one"

    If you change 36 to base 1/6, then you have 36= (1/6)^(-2) so 36^(x+3)= ((1/6)^(-2))(x+3) and so (1/6)^(-2(x+3))= (1/6)^x. Since that function is "one-to-one" (two DIFFERENT values of the exponent can't give the same value), -2(x+3)= x.

    I would have been inclined to use 6 as base (just don't like fractions, I guess). Then 1/6= 6^(-1) so (1/6)^x= 6^(-x). 36= 6^2 so 36^(x+3)= 6^(2(x+3)). Now you have 6^(-x)= 6^(2(x+3)). That means -x= 2(x+3). Of course , those two equations have exactly the same solution.
     
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