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Math Help - Exponential problems

  1. #1
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    Exponential problems

    I think I did these correctly but before I do anymore of the same problems I was wondering if you guys could check them for me


    Simplify with no radicals or negative exponents:


    1. (x(squareroot 16xt^2))^2

    I'm not certain what sign to use on the keyboard for squareroot symbo

    Well this is what I did.

    (x^2)(16xt^2)

    16x^3t^2


    2. (3ab)^-1(a^2b^-1)^2

    (a^2)(b^-2) / (3ab)

    a^2 / 3ab^3

    a / 3b^3


    Any help or advice would be great
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by JonathanEyoon View Post
    I think I did these correctly but before I do anymore of the same problems I was wondering if you guys could check them for me


    Simplify with no radicals or negative exponents:


    1. (x(squareroot 16xt^2))^2

    I'm not certain what sign to use on the keyboard for squareroot symbo

    Well this is what I did.

    (x^2)(16xt^2)

    16x^3t^2
    this looks fine. if you can't use LaTex, use sqrt() for the square root symbol. so you would type [x*sqrt(16x(t^2))]^2


    2. (3ab)^-1(a^2b^-1)^2

    (a^2)(b^-2) / (3ab) ...
    the power you have for a in the numerator is not correct
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    this looks fine. if you can't use LaTex, use sqrt() for the square root symbol. so you would type [x*sqrt(16x(t^2))]^2

    the power you have for a in the numerator is not correct


    Ok the first one is fine =)

    Let me see if I can redo the second one.

    (3ab)^-1((a^2)(b^-1))^2

    From here I will

    (a^4)(b^-2) / (3ab)

    (a^4) / (3ab^3)

    a^3 / 3b^3


    Does that look better?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by JonathanEyoon View Post
    Ok the first one is fine =)

    Let me see if I can redo the second one.

    (3ab)^-1((a^2)(b^-1))^2

    From here I will

    (a^4)(b^-2) / (3ab)

    (a^4) / (3ab^3)

    a^3 / 3b^3


    Does that look better?
    looks good. good job
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  5. #5
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    Quote Originally Posted by Jhevon View Post
    looks good. good job

    Thanks!

    Another question


    This problem

    (e^kt)(e^3)(e) = e^4kt


    Am I right?
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  6. #6
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    Krizalid's Avatar
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    Quote Originally Posted by JonathanEyoon View Post
    This problem

    (e^kt)(e^3)(e) = e^4kt
    So it is e^{kt}\cdot e^3\cdot e

    This requires the power property a^k\cdot a^m\cdot a^n=a^{k+m+n}
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  7. #7
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    Quote Originally Posted by Krizalid View Post
    So it is e^{kt}\cdot e^3\cdot e

    This requires the power property a^k\cdot a^m\cdot a^n=a^{k+m+n}

    Yea that's the problem. Doh I had it wrong then. So instead it should be

    e^3+kt?
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by JonathanEyoon View Post
    Yea that's the problem. Doh I had it wrong then. So instead it should be

    e^3+kt?
    nope. e^{kt} \cdot e^3 \cdot e = e^{kt + 3 + 1}

    remember, e = e^1
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  9. #9
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    Quote Originally Posted by Jhevon View Post
    nope. e^{kt} \cdot e^3 \cdot e = e^{kt + 3 + 1}

    remember, e = e^1


    oopsie forgot about that . Ok my 3rd attempt!!


    So the final answer should be e^4+kt?
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  10. #10
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by JonathanEyoon View Post
    oopsie forgot about that . Ok my 3rd attempt!!


    So the final answer should be e^4+kt?
    yes. but type e^{4 + kt} so that we know the kt is not separate from the e
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  11. #11
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    Quote Originally Posted by Jhevon View Post
    yes. but type e^{4 + kt} so that we know the kt is not separate from the e

    ok!! I always forget to correctly use the parentheses . I'll try to take more notice of them. Thanks alot guys!!
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