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Math Help - SAT math help

  1. #1
    Member OnMyWayToBeAMathProffesor's Avatar
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    SAT math help

    I got these answers right on the practice SAT but can not figure out how to do them properly. Some of them I got right because I guess or just plugged them in. Any help would be wonderful.

    1) If (a+b)^{1/2}=(a-b)^{-1/2} which of the following must be true. The answer is a^2-b^2=1 But how? q.14-423

    2) If 2^{2x}=8^{x-1} what is the value of X? The answer is 3 but how, do you use log or something? q.9-397

    3) The graph above shows Elina's height in inches from age 6 to 12. Elina's height at the age of 12 was what percent higher than her height at the age of 6? Her height at the age of 6 was 45 inches and her height at the age of 12 was 60 inches. The answer is 33 1/3%. But how? q.15-399

    All your help will be wholly and greatly appreciated.
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  2. #2
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    1.
    (a+b)^{1/2} = (a-b)^{-1/2}
    Swtich from fractional index to square root: \sqrt{a+b} = \frac{1}{\sqrt{a-b}}
    Square both sides: a+b = \frac{1}{(a-b)}
    Cross-multiply: (a+b)(a-b) = 1
    Expand (FOIL Method): a^2 - b^2 = 1

    2. 2^{2x} = 8^{x-1}

    Hehe, don't complicate yourself with logs. This is just a SAT question. Basically, what you do is try to get both bases to be the same. We can switch 8 to 2^3, so we get:
    2^{2x} = 2^{3x-3}

    They're equal to each other, so the exponents has to be the same. Thus:
    2x = 3x-3
    x = 3

    3. Percentage increase, very simple. What you do is subtract Elina's height at age of 6 from height at age of 12 and divide by height at age of 6.

    \frac{60-45}{45} = \frac{15}{45} = \frac{1}{3} = 33\frac{1}{3}\%
    Last edited by Chop Suey; July 8th 2008 at 01:15 PM.
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  3. #3
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    Let me add more to my previous post. The SAT is supposed to test your logical reasoning abilities. You gotta be focused, and you also gotta be fast. In problems like these, it's better that you figure out that the cube root of 8 is 2, and thus you can solve this problem faster than doing it with logs. Few seconds difference? Yes, but these few seconds accumulate, and time is a very precious thing in the SAT. Anyways, if you want to do it using logs:

    \log{2^2x}  = \log{8^x-1}
    You can throw the powers outside using the property of logs:
    (2x) \log{2} = (x-1) \log{8}

    2x = (x-1) (\frac{\log{8}}{\log{2}})

    2x = (x-1) (3)

    2x = 3x - 3

    x = 3
    Last edited by Chop Suey; July 8th 2008 at 01:14 PM.
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  4. #4
    Member OnMyWayToBeAMathProffesor's Avatar
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    Thank you very much. Very detailed and excellent explanation.
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