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    Solve the simultaneous equations:

    ab= 25

    log4a - log4b = 3

    How do you do this?
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    Write the second equation as log a - log b = log 64, then write the LHS as log (a/b) and remove the logs. You then have SEs without logs
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    Simultaneously solving Log4a-log4b=3 and ab=25
    Can be written as: log4(a/b)=3
    Removing the log:a/b=43 which is: a/b=64
    Rearranging: a=64b
    Substitute this into the second equation and solve for rest
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    (Original post by teamjudo)
    Simultaneouslysolving Log4a-log4b=3andab=25Can be written as: log4(a/b)=3Removing the log:a/b=43 which is: a/b=64Rearranging: a=64bSubstitute this into the second equation: (64b)b=25Therefore: 64b2=25 So: b2=25/64And: b=sqrt(25/64) which is: b=5/8Again substitute this into the second equationfor a: a(5/8)=25Solve for a: a=25(8/5)Therefore: a=40
    Full solutions (especially when they're horribly formatted) are against forum guidelines. Thanks.
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    (Original post by nerak99)
    Write the second equation as log a - log b = log 64, then write the LHS as log (a/b) and remove the logs. You then have SEs without logs
    Oooh thank you!!

    I've gotten to b2 = 25/64 so would this give two solutions for b ( 5/8 and -5/8) or just one?
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    (Original post by jessyjellytot14)
    Oooh thank you!!

    I've gotten to b2 = 25/64 so would this give two solutions for b ( 5/8 and -5/8) or just one?
    Consider what would happen if b were negative, can you take the logarithm of a negative number?
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    (Original post by Zacken)
    Consider what would happen if b were negative, can you take the logarithm of a negative number?
    Ah okay I guess not then
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    (Original post by jessyjellytot14)
    Ah okay I guess not then
    It's still good practice to write down both answers and then 'reject' one by giving a reason in an A-Level exam, sometimes marks are awarded for this.
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    (Original post by jessyjellytot14)
    Oooh thank you!!

    I've gotten to b2 = 25/64 so would this give two solutions for b ( 5/8 and -5/8) or just one?
    Name:  y=logx.jpg
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    It is not possible to have a log(-x) for example
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    (Original post by teamjudo)
    Name:  y=logx.jpg
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    It is not possible to have a log(-x) for example
    It's quite possible. \log(-x) exists for all real x < 0.
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    (Original post by Zacken)
    It's quite possible. \log(-x) exists for all real x < 0.
    Well if we are noting that --x x>0 is positive then that is a waste of everyones time here.

    But in Core maths, no log of a negative number and so that is a bit of a distraction.

    In Core maths a log can be negative but you cant log a negative
 
 
 
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