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C2 log question...

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    Solve the simultaneous equations:
    Log2(x)+Log8(y)=-1
    Log4(x)+Log2(y)=2

    All logs here are to the base, ie not Log 2x
    Please help
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    Before I go into it, have you got a value of x and y so I can check I'm right?

    Edit:

    I managed to get an answer that works for both of your values. Pretty sure I've never done anything like this for C2 though, what exam board are you with? Also, how many marks was this question?

    To do this question you need to use pretty much all the rules for logs you know. Using log_a x = \frac {log_b x}{log_b a} to give them all the same base and b*log_a x = log_a x^b, log_a x + log_a y = log_a xy and log_a x - log_a y = log_a \frac {x}{y} to simplify the equations.

    I'd advise you have another got at this before you look at my workings. Make sure you put them all in the same base for ease and then manipulate the equations to get values for x and y.

    Spoiler:
    Show

    

log_2 x + log_8 y = -1 = -log_2 2 = log_2 2^{-1}

log_4 x + log_2 y = 2 = 2 \times log_2 2 = log_2 2^2

log_8 y = \frac {log_2 y}{log_2 8} = \frac {log_2 y}{log_2 2^3} = \frac {log_2 y}{3}

log_4 x = \frac {log_2 x}{log_2 4} = \frac {log_2 x}{log_2 2^2} = \frac {log_2 x}{2}

log_2 x + \frac {log_2 y}{3} = log_2 2^{-1}

3 \times log_2 x + log_2 y = 3 \times log_2 2^{-1} = log_2 2^{-3} = log_2 x^3 + log_2 y

((log_2 x) \div 2) + log_2 y = log_2 2^2

log_2 x + 2 \times log_2 y = 2 \times log_2 2^2 = log_2 2^4 = log_2 x + log_2 y^2

log_2 (x^3 \times y) = log_2 2^{-3}

log_2 (x \times y^2) = log_2 2^4

xy^2 = 2^4

x = 2^4 \div y^2

x^3y = 2^{-3}
    

(2^4 \div y^2)^3 \times y = 2^{-3}

2^{12} \div y^6 \times y = 1 \div 8

2^{12} = y^5 \div 8

2^{15} = y^5

y = 8

x \times y^2 = 2^4

64x = 2^4

x = 0.25
  3. Offline

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    you need to do something like replace log2y with 3log8y

    and log2x with 2log4x
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    (Original post by studentccs)
    Solve the simultaneous equations:
    Log2(x)+Log8(y)=-1
    Log4(x)+Log2(y)=2

    All logs here are to the base, ie not Log 2x
    Please help
    You need to use the change of base rule.

     \displaystyle log_2 x + log_8 y = -1 \implies log_2 x + \frac{log_2 y}{log_2 8} = -1 \implies log_2 x + \frac{log_2 y}{3} = -1 \\ \implies x^3 y = \frac18

    In a similar way construct an equation using  log_4 x + log_2 y = 2

    Then sub one equation into another to find the values.

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