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Exponential Equations: Urgent Help

  1. Mar 4, 2007 #1
    I totally need urgent help on these exponential equations. I am so confused write now. I know that the bases for each equation has to be the same and that is x is squared you have to factor or use the quadratic formula to find the solution. I also know that the general steps to solve these equations are to make the bases on each side of the equation the same. You have to set the exponents equal to each other. Lastly, you solve for x. I am still confused, I could do the examples in the lesson, but no these practice problems. Help!!!!!!!!

    1. 9^(2x)=27^(x-1)
    2. 5^(n-1)=1/25
    3. 25^x=5^(x^2-15)
    4. (2^x)(4^(x+5))=4^(2x-1)
    5. (sqrt 3)^(2x+4)=9^(x-2)


    For the first equation, I think that x=-1. For the second equation, I think that n=1/2. For the other equations, I have no idea what to do.
     
    Last edited: Mar 4, 2007
  2. jcsd
  3. Mar 4, 2007 #2

    arildno

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    Set your parentheses right about exponents before expecting help
     
  4. Mar 4, 2007 #3

    cristo

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    Firstly, it would help if you put brackets around the exponents. For example, is question 1. 92x=27x-1 or 92x=27x-1? I presume it is the former, but I disagree with your answer. Show your work; what base would you choose to take on each side? What equation does this yield?
     
  5. Mar 4, 2007 #4
    okay, since the base has to be the same of both sides, I say that 3 would be the base for this equation. I solved the problem and I got -1/5. Is this right?
     
  6. Mar 4, 2007 #5

    arildno

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    Make your equation EXPLICIT at this point!
    Otherwise, we can't help you along..
     
  7. Mar 4, 2007 #6
    1. 9^(2x)=27^(x-1)
    Since the base has to be the same on both sides, I think that the base has to be changed to 3, making the equation read 3^(4)(2x)=3^(3)(x-1). You would set the exponents equal to each other: 8x=3x-1, which would give you x=-1/5. Is this the right answer for this problem, though?
     
  8. Mar 4, 2007 #7

    arildno

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    No, it is not right, not your notation, and not your expression of 9, nor your contraction into an exponent equation.
    We have:
    [tex]9^{2x}=27^{(x-1)}, 9=3^{2},27=3^{3}\to(3^{2})^{2x}=(3^{3})^{x-1}\to3^{2*(2x)}=3^{3*(x-1)}[/tex]
    yielding the exponent equation:
    [tex]2*(2x)=3*(x-1)[/tex]
     
  9. Mar 4, 2007 #8

    D H

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    No. It is easy to verify that [itex]9^{(2x)} \ne 3^{(4(2x)}[/itex]
     
  10. Mar 4, 2007 #9
    So I was right before, x= -1?
     
  11. Mar 4, 2007 #10
    Only if you believe -4=-6.

    Do you understand what's going on here? We're saying that 9 can be represented by 3^2. Therefore, 9^x=(3^2)^x=3^(2x). Now the trick to solving these is to get them to the same base to get an 'exponent' equation.
     
    Last edited: Mar 4, 2007
  12. Mar 4, 2007 #11
    maybe it would help if I showed what I did to get x=-1 and then you could tell me what I am doing wrong.
    9^(2x)=27^(x-1)
    =3^(2)(2x)=3^(3)(x-1)
    =4x=3x-1
    =x=-1
    So where did you get -4=-6 from?
     
  13. Mar 4, 2007 #12

    D H

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    You are guessing (you original post) and being sloppy (your last post). What is [itex]3*(x-1)[/tex]?
     
  14. Mar 4, 2007 #13

    cristo

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    You expand 3(x-1) as 3x-1. This is not correct. It should be 3(x-1)=3x-3
     
  15. Mar 4, 2007 #14

    arildno

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    Do you understand how to multiply out a parenthesis?
     
  16. Mar 4, 2007 #15
    Okay, you're doing quite a few things wrong. Please don't be so liberal with your equality symbols. Break it up. Go the long way. Don't do shortcuts, this is where you're getting messed up.

    9^(2x)=27^(x-1) iff (3^2)^(2x)=(3^3)^(x-1) iff 3^(4x)=3^(3x-3) iff 4x=3x-3 iff x=-3.

    The key thing to realize here is that (a^n)^m=a^(nm). (Where these formulas make sense...of course :| )
     
  17. Mar 4, 2007 #16
    Thank you Zio X. you have been a real help and I get it know.
     
  18. Mar 4, 2007 #17
    Don't thank me, thank arildno. He already told you precisely what I did. Except I gave you a final answer...which I'm kind of regretting.
     
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