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Math Help - Can someone help me solve this logerithm

  1. #1
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    Can someone help me solve this logerithm

    Solve the equation

    (e^4) * e^x2=e^12

    I come up with x^3=12 ,then x^3-12=0.

    I know I'm wrong, but I can't see a way to simplify x^3




    Thanks

    Jason
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    Quote Originally Posted by Darkhrse99 View Post
    Solve the equation

    (e^4) * e^x2=e^12

    I come up with x^3=12 ,then x^3-12=0.

    I know I'm wrong, but I can't see a way to simplify x^3




    Thanks

    Jason
    ?? did you type the question correctly? how do you end up with x^3 anywhere?

    is it supposed to be e^4 e^{x^2} = e^{12}?
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    ?? did you type the question correctly? how do you end up with x^3 anywhere?

    is it supposed to be e^4 e^{x^2} = e^{12}?
    I left out the power of X after (e^4) . I came up with x^3 when i combined x^2 and x. The last part is right. but it's (e^4)^x. It's e the the 4th power times the power of x.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Darkhrse99 View Post
    Yes that is the equation. I came up with x^3 when i combined x^2 and x.
    that's what i'm saying. what x? there is no x here. the only x-term is the x^2

    we have: e^4e^{x^2} = e^{x^2 + 4} = e^{12}

    now equate the powers and continue
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  5. #5
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    Quote Originally Posted by Jhevon View Post
    that's what i'm saying. what x? there is no x here. the only x-term is the x^2

    we have: e^4e^{x^2} = e^{x^2 + 4} = e^{12}

    now equate the powers and continue
    It's (e^4)^x * e^x^2=e^12
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Darkhrse99 View Post
    It's (e^4)^x * e^x^2=e^12
    ok, one more time for clarity's sake. it is \left( e^4 \right)^x e^{x^2} = e^{12}?

    read carefully this time and make sure this is the problem
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  7. #7
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    Quote Originally Posted by Jhevon View Post
    ok, one more time for clarity's sake. it is \left( e^4 \right)^x e^{x^2} = e^{12}?

    read carefully this time and make sure this is the problem
    Yes that is the correct problem. Sorry about the confusion. I was editing while you were posting.
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Darkhrse99 View Post
    Yes that is the correct problem. Sorry about the confusion. I was editing while you were posting.
    okay, so you should know two things here, x^3 will still not come into play:

    \left( x^a \right)^b = x^{ab} ....................(1)

    x^a \cdot x^b = x^{a + b} ................(2)


    now on to the problem

    \left( e^4 \right)^x e^{x^2} = e^{12}

    \Rightarrow e^{4x}e^{x^2} = e^{12} ................by (1)

    \Rightarrow e^{x^2 + 4x} = e^{12} ................by (2)

    equating the powers we get

    x^2 + 4x = 12

    now conitnue
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  9. #9
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    I get

    x^2+4x=12
    x^2+4x-12=0
    (x-2) (x+6)=0
    x=2 x=-6 Correct?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Darkhrse99 View Post
    I get

    x^2+4x=12
    x^2+4x-12=0
    (x+4) (x-3)=0
    x=-4 x=3 Correct?
    no

    expand (x + 4)(x - 3), what do you get?

    i guarantee you do not get x^2 + 4x - 12
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  11. #11
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    x=-6 x=2 My hands aren't following my brain tonight.uugghh
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  12. #12
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Darkhrse99 View Post
    x=-6 x=2 My hands aren't following my brain tonight.uugghh
    yes, those are the solutions to the quadratic. make sure to check both solutions in the original equation to make sure they work
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  13. #13
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    Ok thanks for the help and sorry for the confusion. What program are you guys workin with to write out problems.?
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  14. #14
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Darkhrse99 View Post
    Ok thanks for the help and sorry for the confusion.
    you're welcome

    and that's ok. good luck with your class
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