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Math Help - Questions about exponents and absolute values

  1. #1
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    Questions about exponents and absolute values

    Ok the first one is: 7^2x+3 = 5^3x-2

    I Understand that I can put them in log form, (2x+3)log 7 = (5x-2)log5
    but, my teacher mentioned we should use the natural log? So how do I go about doing so?

    2. 8^x2 * 4^-3x = 1/2
    How I did it:
    2^(3x^2) * 2^-6x = 1/2
    = 3x^2 * -6x = 1/2
    = -18x^3 = 1/2

    Some how that doesn't seem right to me.
    I thought I could make the bases equal, so I used 2 for both the 8 and the 4. is this correct? does the 1/2 have to have to the same base also?

    3. 8-2|6-2x| ≤ -4
    -8 -8
    = -2|6-2x| ≤ -12

    then I divided by 2 on both sides to get |6-2x| ≤ -12

    Is that right? should I be dividing by -2 or adding?

    4. Last one:
    2^x+1 * 8^-x = 4

    Same as the first one, I set the bases as 2, So i get
    2^x+1 * 2^-3x = 2^2
    Should i be doing that for the 4 also? it doesn't seem right again...Thx in advance for the help.
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  2. #2
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    Quote Originally Posted by ktpinnock View Post
    Ok the first one is: 7^2x+3 = 5^3x-2

    I Understand that I can put them in log form, (2x+3)log 7 = (5x-2)log5
    but, my teacher mentioned we should use the natural log? So how do I go about doing so?

    doesn't matter ... same a log base 10.

    (2x+3)ln 7 = (3x-2)ln 5


    solve for x

    2. 8^x2 * 4^-3x = 1/2
    How I did it:
    2^(3x^2) * 2^-6x = 1/2

    2^(3x^2-6x) = 2^(-1)

    3x^2 - 6x = -1


    3. 8-2|6-2x| ≤ -4
    -8 -8
    -2|6-2x| ≤ -12

    divide both sides by -2 ... dividing by a negative value changes the direction of the inequality.

    |6-2x| > 6

    4. Last one:
    2^x+1 * 8^-x = 4

    2^(x+1) * 2^(-3x) = 2^2

    2^(1-2x) = 2^2

    exponents are equal ...

    ...
    Last edited by skeeter; September 6th 2009 at 06:06 PM. Reason: fixed exponent in last problem
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    o0o0o, thanks a lot for the quick response and clearing the mist that was forming over my brain

    One more quick question...
    (4 is the base)
    log4 (x) + log 4 (x-4) = 1, Solve for x:

    I did log4 (x)(x-4) = 1

    i get confused after that because my teacher said it should be set = 4..which I don't get, can you please explain? or anyone for that matter...thanks again
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  4. #4
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    Quote Originally Posted by ktpinnock View Post
    o0o0o, thanks a lot for the quick response and clearing the mist that was forming over my brain

    One more quick question...
    (4 is the base)
    log4 (x) + log 4 (x-4) = 1, Solve for x:

    I did log4 (x)(x-4) = 1

    i get confused after that because my teacher said it should be set = 4..which I don't get, can you please explain? or anyone for that matter...thanks again
    one method ...

    \log_4[x(x-4)] = 1

    \log_4[x(x-4)] = \log_4{4}

    x(x-4) = 4


    another ...

    \log_4[x(x-4)] = 1

    change to an exponential ...

    4^1 = x(x-4)


    now solve ... don't forget to check any solution(s) in the original equation.
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  5. #5
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    ooo ok, I was on the right track...thanks for the directions..I see ur also a fellow spammer of ellipses

    Edit: Last one I swear D:

    log(7x-12) = 2logx

    Can i just drop the log since they are the same base?

    and write 7x-12 = x^2?
    Last edited by ktpinnock; September 6th 2009 at 06:55 PM.
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  6. #6
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    Quote Originally Posted by ktpinnock View Post
    ooo ok, I was on the right track...thanks for the directions..I see ur also a fellow spammer of ellipses
    correction, ... ellipsis


    ellipses ...

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  7. #7
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    Quote Originally Posted by skeeter View Post
    correction, ... ellipsis


    ellipses ...

    Actually ellipses is also the plural of ellipsis, which is what i was going for...but i guess this IS a math forum...
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  8. #8
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    Quote Originally Posted by ktpinnock View Post
    Actually ellipses is also the plural of ellipsis ...
    ... and you learn something new every day.
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    Quote Originally Posted by skeeter View Post
    ... and you learn something new every day.
    yea aha...well since your still here...


    log(7x-12) = 2logx

    Can i just drop the log since they are the same base?

    and write 7x-12 = x^2?

    and Factoring..

    64a^6 + 27b^3
    i did (8a^3) (8a^3) + (3b)(3b)(3b)

    and for 2x^4-162
    2(x^4 - 81)
    2(x^2 - 9) (x^2 +9)
    2(x - 3) (x+3) (x^2 - 9)

    Right? (doubting they are tho)
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  10. #10
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    Quote Originally Posted by ktpinnock View Post
    yea aha...well since your still here...


    log(7x-12) = 2logx

    Can i just drop the log since they are the same base?

    and write 7x-12 = x^2? yes

    and Factoring..

    64a^6 + 27b^3
    i did (8a^3) (8a^3) + (3b)(3b)(3b)

    (4a^2)^3 + (3b)^3

    factoring pattern is x^3 + y^3 = (x + y)(x^2 - xy + y^2)

    and for 2x^4-162
    2(x^4 - 81)
    2(x^2 - 9) (x^2 +9)
    2(x - 3) (x+3) (x^2 + 9)
    btw ... start new problems with a new thread.
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  11. #11
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    Oh okay, I figured I'd just do it here instead of wasting thread space, but I'll do that next time, and thanks for being patient with my slowness.
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