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Solving Exponential Equations

  1. Aug 21, 2008 #1
    1. The problem statement, all variables and given/known data
    Solve the following:
    Question 1:

    [tex]\sqrt[5]{256}\div\sqrt[6]{64}=2^{x}[/tex]

    Question 2:

    [tex]\frac{(9^{2x-1})^{3}(3^{3x})^{2}}{(27^{x+2})^{4}}=81^{3}[/tex]

    2. Relevant equations
    None that I am aware of.

    3. The attempt at a solution
    Question 1:

    [tex]\sqrt[5]{2^{8}}\div2^{1}=2^{x}[/tex]

    Question 2:

    [tex]\frac{(3^{4x-2})^{3}(3^{6x})}{(3^{3x+6})^{4}}=3^{12}[/tex]

    [tex]3^{12x-6+6x-(12x+24)}=3^{12}[/tex]

    [tex]6x=39[/tex]

    [tex]x=6.5[/tex]

    Sorry that the attempt for question 1 was lacking so much, I was just unsure of what to do with the square roots etc...As for number 2, the answer I got is incorrect, when it says that the answer in the back of my text book is 7.

    Answer Key:
    Question 1:
    [tex]\frac{3}{5}[/tex]
    Question 2:
    7

    Thanks to everyone who helps.
     
  2. jcsd
  3. Aug 21, 2008 #2

    nicksauce

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    For #1 you almost have it... just express the fifth root of 2^8 as 2 raised to some power, then use the fact that 2^a / 2^b = 2^(a-b).

    For #2 just check your numbers again. -6 -24 = -30. 12 + 30=42.
     
  4. Aug 22, 2008 #3
    1)

    Two things that is helpful to know:

    [tex]\sqrt[a]{n^b} = n^{ \frac{b}{a} } [/tex]

    and

    [tex] \lg{\frac{a}{b}} = \lg{a} - lg{b} [/tex]


    You don't need the logarithm in this case though, since all your numbers are factors of two.

    k
     
  5. Aug 22, 2008 #4
    Ok thanks, I realized what I had done wrong for number two earlier today but I didn't know the rule about number one. I'm just starting into the logarithms so things should start becoming more familiar as well. Thank you.
     
  6. Aug 23, 2008 #5

    HallsofIvy

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    Well, how about these:
    [tex]^n\sqrt{a}= a^\frac{1}{n}[/tex]
    [tex](a^n)^m= a^{mn}[/tex]
    [tex]a^m\div a^n= a^{m-n}[/tex]

     
  7. Aug 23, 2008 #6
    Yes sorry I should've added those (the ones I used). I have another question though, I don't know what to do when adding two variables with exponents, for example what would something like this become:

    [tex]x^{4}+x^{3}=[/tex]

    or

    [tex]x^{4}-x^{7}=[/tex]

    I couldn't find it on the internet and I don't remember what to do in this case...
     
  8. Aug 23, 2008 #7

    HallsofIvy

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    There is no rule for adding different powers of the same base. That is exactly why polynomials are written as they are.
     
  9. Aug 23, 2008 #8
    Ok I've tried so many things and I can't seem to figure out these two questions. Here they are:

    Question 3:

    [tex]2^{x-1}-2^{x}=2^{-3}[/tex]

    Question 4:

    [tex]3^{x+1}+3^{x}=36[/tex]

    My work is a mess as I have tried so many variations which are probably way off course.
     
  10. Aug 23, 2008 #9

    nicksauce

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    you can write 2^(x-1) as 2^x*2^(-1)
    Thus you can factor the LHS, and isolate 2^x. Question 4 should be similar.
     
  11. Aug 23, 2008 #10
    Can you explain what you mean? So it would become:

    [tex]2^{x}\times2^{-1}-2^{x}=2^{-3}[/tex]

    But I can't really combine the like terms ([tex]2^{x}[/tex]), can I? Because isn't the first one considered "attached" to the second one? ([tex]2^{x} and 2^{-1}[/tex])

    And for the second question:

    [tex]3^{x}\times3^{1}+3^{x}=36[/tex]

    [tex]6^{x}\times3=36[/tex]

    [tex]18^{x}=36[/tex]

    I'm not sure what to do from here...maybe I'm already wrong?
     
  12. Aug 24, 2008 #11

    nicksauce

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    [tex]
    2^{x}\times2^{-1}-2^{x}=2^{-3}
    [/tex]

    You can write this as
    [tex]
    2^x(2^{-1}-1})=2^{-3}
    [/tex]

    Now write the term in the parentheses as 2 to some power, and you should be all set.
     
  13. Aug 24, 2008 #12
    So it's:

    [tex]2^{x}(-2^{-1})=2^{3}[/tex]

    But seeing as the answer in the back says no solution, I'm assuming it's not possible to solve this with a negative base seeing as the rest are positive? (2 and -2)
     
  14. Aug 24, 2008 #13

    nicksauce

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    Right. You have 2^x = (some negative number). There is no value of x for which this is true, as 2^x is always positive. (I should have noticed this earlier, but didn't think of it for what ever reason)
     
  15. Aug 24, 2008 #14
    Oh ok that makes sense, with the other question (question 4) I know how to go about it now but how do I establish a common base between 3 and 36?
     
  16. Aug 24, 2008 #15
    Actually I'll just show my work for now so you can see the progress I've made so far:

    [tex]3^{x+1}+3^{x}=36[/tex]

    [tex]3^{x}(3^{1}+1)=36[/tex]

    Ok nevermind I thought I had an idea but I'll leave this up for reference.
     
  17. Aug 24, 2008 #16

    HallsofIvy

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    And 3+1= ?

     
  18. Aug 24, 2008 #17
    [tex]3^{x}(4)=36[/tex]

    Well its obvious that x is 2 but how do I do that algebraically...I can't go [tex]12^{x}=36[/tex] because x=2 wouldnt work...This probably looks like such a dumb question I just can't figure out how to get a common base...Can I go:

    [tex]3^{x}(4)=3^{2}(4)[/tex]

    [tex]x=2[/tex]

    Does that make sense?
     
  19. Aug 24, 2008 #18

    nicksauce

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    Yes that makes sense.
     
  20. Aug 24, 2008 #19
    Oh ok thank you, I just didn't know you could have a base with two separate numbers.
     
  21. Aug 25, 2008 #20
    leave the left hand side as it was (from the start) change right hand side to
    [tex]3*3^{2}[/tex] note this is equal to 36
    now all the bases are the same and they can cancel/disapear (index laws)
    so now you have
    (x+1)+x=1+2
    and i'm sure you can go from here
     
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