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    After spending all day on this chapter, these are the questions I can't find the answers to after multiple rechecking.

    I'd appreciate any help at all.

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    (2D) (5) (a) (i)


    *more to come*
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    (2E) (2) (e) (ii)

    Need to express in terms of x, y and z

    When
    X = log a
    Y = log b
    Z = log c
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    2E 3 e i

    Need to find x
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    2G 1 d i
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    The two I got wrong.
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    what are white questions?
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    (Original post by TeeEm)
    what are white questions?
    The questions in my book are colour coded as white, yellow, green, blue and red
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    (Original post by IFoundWonderland)
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    (2D) (5) (a) (i)
    Why have you crossed out 2^32? It's correct.
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    (Original post by IFoundWonderland)
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    2E 3 e i

    Need to find x
    You said \log_3 (\text{expanded quadratic}) = 2, so to get rid of the log, you need to do 3^2, not 2^3.
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    (Original post by IFoundWonderland)
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    2G 1 d i
    When you divide by 3, at the very end (3x = blah, to x = blah/3), you need to multiply the entire denominator by 3. That is:

    \displaystyle \frac{\log_3 7 - 3}{\log_3 2 - 1} \times \frac{1}{3} = \frac{\log_3 7 - 3}{3(\log_3 2 - 1)} = \frac{\log_3 7 - 3}{3\log_3 2 - 3 }
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    (Original post by IFoundWonderland)
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    The two I got wrong.
    You have \displaystyle 2x \ln 3 = \ln 5, which I agree with, now divide both sides by \ln 3 to get:

    \displaystyle \frac{2x \ln 3}{\ln 3} = \frac{\ln 5}{\ln 3} \iff 2x = \frac{\ln 5}{\ln 3}

    You seem to have done: \displaystyle 2x \ln 3 = \ln 5 \iff 2x = \frac{\ln 3}{\ln 5}, but that's not true/correct, can you see why?
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    (Original post by IFoundWonderland)
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    The two I got wrong.
    Whilst we have the rule that \frac{a +b}{c} = \frac{a}{b} + \frac{a}{c}, it is not true that \frac{a}{b+c} = \frac{a}{b} + \frac{a}{c}! Test this out for yourself: \frac{1}{1+1} \neq \frac{1}{1} + \frac{1}{1} = 2 \neq \frac{1}{1+1} = \frac{1}{2}.

    So whilst your \displaystyle \frac{\ln 2}{\ln 5 - \ln 2} is correct, that's your final answer and you should leave it as such!
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    (Original post by IFoundWonderland)
    The questions in my book are colour coded as white, yellow, green, blue and red
    Have I missed anything?
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    Thank you so much for your patience and help :woo:




    (Original post by IFoundWonderland)
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    (2E) (2) (e) (ii)

    Need to express in terms of x, y and z

    When
    X = log a
    Y = log b
    Z = log c
    (Original post by Zacken)
    Have I missed anything?
    Just this one

    (Original post by Zacken)
    Why have you crossed out 2^32? It's correct.
    The book says it's wrong :dontknow:

    Also, for this one, I corrected where you said, but still got the wrong answer :/ I got a negative but the answer wants a positive (I'm guessing 2 should be negative and 8 should be positive but idk how to get this)
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    (Original post by IFoundWonderland)
    Thank you so much for your patience and help :woo:
    No problemo! Give me a sec, just wrapping some stuff up and I'll answer the rest of your questions in ~30 minutes!
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    (Original post by IFoundWonderland)
    Also, for this one, I corrected where you said, but still got the wrong answer :/ I got a negative but the answer wants a positive (I'm guessing 2 should be negative and 8 should be positive but idk how to get this)
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    You have x^2 - 6x -16, but you've written: x^2 - 2x + 8x -16, note the two middle terms are -2x + 8x = +6x. you need it to be -6x, that is, you need:

    \displaystyle 

\begin{equation*}x^2 + 2x - 8x - 16\end{equation*}

    Pulling out -1 as a common factor in the last two terms:

    \displaystyle 

\begin{equation*}x^2 + 2x - (8x + 16) \end{equation*}

    Pulling out 8 as a common factor from the brackets (note that you could have done the above step and this one simultaneously by pulling out -8)

    \displaystyle 

\begin{equation*}x(x+2) - (8(x+2)) = x(x+2) - 8(x+2) = \cdots \end{equation*} can you take it from here?
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    (Original post by Zacken)
    You have x^2 - 6x -16, but you've written: x^2 - 2x + 8x -16, note the two middle terms are -2x + 8x = +6x. you need it to be -6x, that is, you need:

    \displaystyle x^2 + 2x - 8x - 16 = x^2 + 2x - (8x + 16)  \\ = x(x+2) - (8(x+2)) = x(x+2) - 8(x+2) = \cdots can you take it from here?
    Ahh yes, of course :facepalm:

    Thank you so much for all your help - honestly don't have the words to express my gratitude.

    You're an absolute superstar
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    (Original post by IFoundWonderland)
    Ahh yes, of course :facepalm:

    Thank you so much for all your help - honestly don't have the words to express my gratitude.

    You're an absolute superstar
    Edited my post to make it cleared!

    Don't thank me just yet! I've got two more of your questions to reply to.

    Protip: if you can't seem to factorise something correctly, it's worth just using the quadratic equation to find the roots in an exam or something.
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    (Original post by IFoundWonderland)

    The book says it's wrong :dontknow:
    You're correct, the book is wrong!
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    (Original post by IFoundWonderland)
    Just this one
    Aaah, you're almost very nearly there! Note that you have:

    \displaystyle 

\begin{equation*}\log 5 + y + \log 2 + 2z\end{equation*}

    Shiftin things around:

    \displaystyle 

\begin{equation*}\log 5 + \log 2 + y + 2z\end{equation*}

    Using the addition -> multiplication rule thingy for logs.

    \displaystyle

\begin{equation*}\log (5\times 2) + y + 2z = \log 10 + y + 2z\end{equation*}

    Using the fact that \log 10 = \log_{10} 10 = 1 (in this case):

    \displaystyle 

\begin{equation*}1 + y + 2z\end{equation*}

    Note that we've used the rule \log_a b + \log_a c = \log_a bc with a=10, b=5 and c=2.
 
 
 
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