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    x + y = 35

    log_2_X - log_4_Y = 1


    Solve the simultaneous equations
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    rearrange: x=35-y
    substitute in to the second equation:

    log2(35-y) - log4y = 1
    using log rules, this is the same as
    log ((70-2y)/4y) = 1
    We know that log10=1
    So that means that (70-2y)/4y=10
    70-2y=40y
    70=42y
    y=70/42=5/3
    x=35-y=33 1/3
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    Hey i am an idiot, please can you explain this part to me, which rule, and how u did it
    ever so grateful!

    log2(35-y) - log4y = 1
    using log rules, this is the same as
    log ((70-2y)/4y) =
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    Well.. I got y = 25 x = 10 :/
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    I used the rule log a - log b = log(a/b)

    Narin: have you tried putting that back into the equation and testing it? Because it didn't work when I tried your answer, and it worked with mine..
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    i dont understand what rule u used?
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    There is a rule that if you add any two logs with the same base (by writing just log with no base number, it is assumed it is base 10), then you can make it into one log, if you multiply the numbers being logged. So log a + log b = log ab

    This rule extends that if you subtract one log from another, you divide the numbers:
    log a - log b = log (a/b)
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    (Original post by naystar)
    I used the rule log a - log b = log(a/b)

    Narin: have you tried putting that back into the equation and testing it? Because it didn't work when I tried your answer, and it worked with mine..
    Well you see when I tried with mine it works, when i try with yours, it doesn't.
    I think it comes down to one of us using a rule wrongly, since calculators don't work in base 2.

    With my answer:
    X = 10
    log_2_10 = log 10 / log 2 = 3.32

    Y = 25

    Log_4_25 = log 25 / log 4 = 2.32

    3.32 - 2.32 = 1
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    can anyone cofirm the correct answer?
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    Is this log base 2? I may have misunderstood how you wrote it out. Thugzmansion can you clarify what base each log was? (as in can you 'go advanced' and write out the question with super and subscripts so we can tell what the question is).

    If the logs are of base 2 and 4, then yes, my answer would be wrong. I may have misunderstood how they were written out.
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    I got,

    log_4_y
    = (log_2_y)/(log_2_4)
    = (log_2_y)/2

    So
    log_2_x - (log_2_y)/2 = 1
    2log_2_x - log_2_y = 2
    log_2_x^2 - log_2_y = 2
    log_2_(x^2/y) = log_2_4
    x^2/y = 4
    x^2 = 4y

    Substituting x + y = 35, so y = 35 - x into the equation,

    x^2 = 4 (35 - x)
    x^2 = 140 - 4x
    x^2 + 4x - 140 = 0
    x = 10 or -14
    Since there is no value for log_x_(-14), x = 10

    And therefore I got x = 10, y = 25 as well
 
 
 
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