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    11) a) Show that log_43=log_2(\sqrt{3})
    b) Hence or otherwise solve the simultaneous equations:
    2log_2y=log_43+log_2x
    3^y=9^x
    x,y > 0

    I've done part a, but not sure how to start part b.
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    (Original post by ViralRiver)
    11) a) Show that log_43=log_2(\sqrt{3})
    b) Hence or otherwise solve the simultaneous equations:
    2log_2y=log_43+log_2x
    3^y=9^x
    x,y > 0

    I've done part a, but not sure how to start part b.
    I'd start with your second equation. From what you know of the laws of indices you should be able to get y in terms of x.
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    I can derive ylog3=xlog9?
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    (Original post by ViralRiver)
    I can derive ylog3=xlog9?
    Well, what's 9 in terms of 3?
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    3^2, but I don't see how that helps >< .

    OOOHH!! y = 2x
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    (Original post by ViralRiver)
    3^2, but I don't see how that helps >< .

    OOOHH!! y = 2x
    Grin! Now sub into your first equation and use the hint from a)
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    Ok, so I now have 2log_2(2x)-log_2(x)=log_2(\sqrt{3}) I'm not sure how to put everything in terms of x though.
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    (Original post by ViralRiver)
    Ok, so I now have 2log_2(2x)-log_2(x)=log_2(\sqrt{3}) I'm not sure how to put everything in terms of x though.
    Using the laws of logs, try and combine the left hand side into one log function.

    a log x = log (x^a)

    log b - log a = log (b/a)
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    (Original post by ViralRiver)
    Ok, so I now have 2log_2(2x)-log_2(x)=log_2(\sqrt{3}) I'm not sure how to put everything in terms of x though.
    do what the person above just said but be careful with the 2 in front of the first term, youll have to take it up as a power
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    Ahh ok, doing that I have x=\frac{\sqrt{3}}{4}?
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    (Original post by ViralRiver)
    Ahh ok, doing that I have x=\frac{\sqrt{3}}{4}?

    Why the question mark. Have faith, you got it right. You just need y now.
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    Ahh yes, got y and they're both right. I'm just stuck on the last question now.

    12). a) Given that 3 + 2log_2x=log_2y, show that y=8x^2.

    Do I need to turn 3 into a logarithm, or is that unnecessary?
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    (Original post by ViralRiver)
    Ahh yes, got y and they're both right. I'm just stuck on the last question now.

    12). a) Given that 3 + 2log_2x=log_2y, show that y=8x^2.

    Do I need to turn 3 into a logarithm, or is that unnecessary?
    There's no simple yes/no answer to that one. Do it, and see where you get (it will work it).
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    That's the thing, I'm not sure how to turn it into one >< . Because 3 \not= log3.
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    (Original post by ViralRiver)
    That's the thing, I'm not sure how to turn it into one >< . Because 3 \not= log3.

    You can write 3 as 3\times\log_22 since \log_22 is equal to 1. And use that as the basis for conversion, using the rules previously stated.
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    Thanks ! I've just worked out parts b c and d from that so I must be getting the hang of it :P .
 
 
 
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