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Exponential Eqn

  1. Dec 6, 2004 #1
    To solve this exponential equation (4)(2)^5x=square root of 2
    I know the common base is 2 but how do i make the square root of 2 also a common base? Do i multiply both sides by ^2? When I did this I got x=1/40 im not sure if this is the right way to do this question can someone tell me :confused:
     
  2. jcsd
  3. Dec 6, 2004 #2
    [tex]4*(\sqrt{2} * \sqrt{2})^{5x} = \sqrt{2} [/tex]
     
  4. Dec 6, 2004 #3
    I think you might be on the right track... could you explain what you did? I might have done it the way you were thinking but I got a different answer.
     
  5. Dec 6, 2004 #4
    What I did

    well first I found the common base of left side and got
    (2^2) (2)^5x=square root of 2
    then to get rid of the square root on the right I wasnt sure what to do but I decided to multiply both sides by ^2 and got [(2^2) (2)^5x]^2=2 so the exponents were [(2)(5x)]2 = 2 and then when i further simplified I got the exponent x to =1/40
     
  6. Dec 6, 2004 #5
    When you got to here,
    , what did you do with the ^2 after the 2?
     
  7. Dec 6, 2004 #6
    I am not sure what u are talking about. :uhh:
     
  8. Dec 6, 2004 #7

    Gokul43201

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    Write both sides as powers of sqrt(2) and then compare powers.
     
  9. Dec 6, 2004 #8
    lol sorry still confused can someone show me :uhh:
     
  10. Dec 7, 2004 #9

    shmoe

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    From here "[(2^2) (2)^5x]^2=2"

    to here

    "so the exponents were [(2)(5x)]2 = 2"

    You ran into some difficulty. The first equation is [tex]\left[2^{2}2^{5x}\right]^2=2[/tex], which is fine. When you combined the terms inside the [] you multiplied the exponents when you should have added them. You should get [tex]\left[2^{2+5x}\right]^2=2[/tex]. Can you manage from here?
     
  11. Dec 7, 2004 #10
    Ok I see my mistake but not sure if my answer is right now i got x=-1/5
     
  12. Dec 7, 2004 #11

    Gokul43201

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    No, but I think I see where you're going wrong.
    Does this help (I've made a small change) ? [tex]\left[2^{2+5x}\right]^2=2^1[/tex].
     
  13. Dec 7, 2004 #12

    shmoe

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    Hold on-remember the power of 2 outside the square brackets [].

    It's always a good idea after solving one of these to test your solution back in the original equation.
     
  14. Dec 7, 2004 #13
    :confused: ok this is what I am doing now the exponents are (2+5x)^2=1 this to 4+10x=1 then i get -3/10 =x

    oh to check myself where do i sub in x? which part?
     
  15. Dec 7, 2004 #14

    shmoe

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    Your original equation was [tex]4*(2)^{5x} = \sqrt{2} [/tex]. If you replace the "x" in this equation with "-3/10", the equation should be true.
     
  16. Dec 7, 2004 #15
    :smile: :rofl: YAYAYA
    IM SOO HAPPY LOL I finally got that THANKS SO MUCH everyone ESP SHMOE
     
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