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Thread: (UK) A-level Maths H/W.

  1. #1
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    Thumbs down (UK) A-level Maths H/W.

    Hi, i used this forum for my GCSE mathematics, and I managed to get an A in the end! Firstly, if any of you remember my panicked questions, i would like to thank you for all the help given as similar questions came up in the test requiring that method to solve them! I moved to higher tier 4 weeks before i broke up from school, so had to ask friends and you helpful people for help, but thankfully it was worth it and i got the A.. so thank you all who helped!

    Anyway.. I have moved on. I am now doing A2 level maths along with biology, chemistry, english language (and a spanish course for beginners). My teacher has given me some terribly hard questions and i was wondering if anyone could get me going on these.. the whole class thinks they are hard but i need to try and answer some! So hear goes..

    1a) Given that 2^a = 8^b, express a in terms of b.
    1b) State the value of x for which 9^x = 3.
    2) If x^p = (√x)^3, x^q = 1/x^2 and x^r=x^p/x^q
    a) Find the value of x
    b) Evaluate x^r when x = 4
    3a) Express (1+√2)^2 in the form a+b√2, where a and b are integers to be found.
    3b) By substituting x = y^1/2, or otherwise, solve the equation 6y^1/2 = y+9

    There are more questions.. many more but to get me started some help on those would be great
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Danielisew View Post
    1a) Given that 2^a = 8^b, express a in terms of b.
    Do you remember the properties of logarithms?
    $\displaystyle 2^a = 8^b$

    $\displaystyle 2^a = (2^3)^b$

    $\displaystyle 2^a = 2^{3b}$

    $\displaystyle log_2(2^a) = log_2(2^{3b})$

    $\displaystyle a = 3b$

    Quote Originally Posted by Danielisew View Post
    1b) State the value of x for which 9^x = 3.
    $\displaystyle 9^x = 3$

    $\displaystyle (3^2)^x = 3^1$

    $\displaystyle 3^{2x} = 3^1$

    $\displaystyle log_3(3^{2x}) = log_3(3^1)$

    $\displaystyle 2x = 1$
    etc.

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Danielisew View Post
    2) If x^p = (√x)^3, x^q = 1/x^2 and x^r=x^p/x^q
    a) Find the value of x
    b) Evaluate x^r when x = 4
    Try to do these the same way I did the last one. You will need to recall that
    $\displaystyle \sqrt{x} = x^{1/2}$

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Danielisew View Post
    3a) Express (1+√2)^2 in the form a+b√2, where a and b are integers to be found.
    You should be able to do this one!
    $\displaystyle (1 + \sqrt{2} )^2 = (1 + \sqrt{2} )(1 + \sqrt{2} )$

    Now FOIL it out:
    $\displaystyle = 1 \cdot 1 + 2 \sqrt{2} + \sqrt{2} \cdot \sqrt{2} = $?

    Quote Originally Posted by Danielisew View Post
    3b) By substituting x = y^1/2, or otherwise, solve the equation 6y^1/2 = y+9
    You want to rewrite this in the form of a quadratic equation:
    $\displaystyle 6y^{1/2} = y + 9$

    $\displaystyle y - 6y^{1/2} + 9 = 0$

    Now let $\displaystyle z = y^{1/2}$, so $\displaystyle z^2 = y$, giving:
    $\displaystyle z^2 - 6z + 9 = 0$

    Factor this and solve it. Don't forget to take the square root of z at the end to get your y.

    -Dan
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  5. #5
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    Hello, Danielisew!

    For most of these, you're expected to know that:

    . . If $\displaystyle b^x\:=\:b^y$, then $\displaystyle x = y.$

    Given two equal exponentials with equal bases, their exponents are equal.


    1a) Given that $\displaystyle 2^a \,= \,8^b$, express $\displaystyle a$ in terms of $\displaystyle b.$
    We have: .$\displaystyle 2^a\:=\:(2^3)^b\quad\Rightarrow\quad 2^a\:=\:2^{3b}$

    Therefore: .$\displaystyle \boxed{a \:=\:3b}$



    1b) State the value of $\displaystyle x$ for which $\displaystyle 9^x \,= \,3.$
    We have: .$\displaystyle (3^2)^x\:=\:3\quad\Rightarrow\quad3^{2x} \:=\:3^1$

    Therefore: .$\displaystyle 2x \:=\:1\quad\Rightarrow\quad\boxed{ x \:=\:\frac{1}{2}}$



    2) If $\displaystyle x^p = (\sqrt{x})^3,\;x^q = \frac{1}{x^2}\;\text{and }x^r = \frac{x^p}{x^q}$

    a) Find the value of $\displaystyle x$ . . . . Did you mean $\displaystyle {\color{blue}r}$ ?
    $\displaystyle x^p \:=\:(\sqrt{x})^3\quad\Rightarrow\quad x^p \:=\:x^{\frac{3}{2}}\quad\Rightarrow\quad p \:=\:\frac{3}{2}$

    $\displaystyle x^q \:=\:\frac{1}{x^2}\quad\Rightarrow\quad x^q \:=\:x^{-2}\quad\Rightarrow\quad q \:=\:-2$

    Then: .$\displaystyle x^r\:=\:\frac{x^{\frac{3}{2}}}{x^{-2}}\quad\Rightarrow\quad x^r\:=\:x^{\frac{7}{2}}\quad\Rightarrow\quad\boxed { r \:=\:\frac{7}{2}}$


    b) Evaluate $\displaystyle x^r$ when $\displaystyle x = 4$
    $\displaystyle \text{If }x = 4\!:\;\;x^r \;=\;4^{\frac{7}{2}} \;=\;\boxed{128}$

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