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    Solve the equation: 2log5 (x) - log5 (3x) = 1

    am i supposed to factorise the x?
    x[2log5(1) - log5 (3)] = 1 ?

    or do i have to use the log laws? i'm confused because either way it doesn't seem to work
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    Use log laws to manipulate the LHS so that you have \log_5(\text{something})=1.
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    (Original post by sadboynerd)
    Solve the equation: 2log5 (x) - log5 (3x) = 1

    am i supposed to factorise the x?
    x[2log5(1) - log5 (3)] = 1 ?

    or do i have to use the log laws? i'm confused because either way it doesn't seem to work
    You can't do that, by factoring out a  x , you're implying that  \log (x) = x\log(1) which is not the case.  \log() is a function, whatever goes inside the brackets is called the 'argument' of the function.

    To prove to yourself that  \log(a) \not= a\log(1) , try some values for  a .

    Example: Taking  a = 5 , by your definition

     \log 5 = 5\log1

     \log 5  \approx 0.6987...

     5\log1 = 0

     \therefore \log 5 \not= 5\log1
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    (Original post by sadboynerd)
    Solve the equation: 2log5 (x) - log5 (3x) = 1

    am i supposed to factorise the x?
    x[2log5(1) - log5 (3)] = 1 ?

    or do i have to use the log laws? i'm confused because either way it doesn't seem to work
    are you sure it's not 2log5(x)-log5(3x)=-1? I got a solution but it only works for -1

    UPDATE: false alarm. I made a mistake. I found the solution. You need to rewrite the 1 as log5(5), move it to the other side of the equation to make it equal 0, and work from there. Feel free to ask in case any more help is needed.
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    (Original post by tim_72)
    are you sure it's not 2log5(x)-log5(3x)=-1? I got a solution but it only works for -1
    there is a solution to OPs equation, an integer one at that
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    (Original post by DylanJ42)
    there is a solution to OPs equation, an integer one at that
    yeah i made a mistake, i edited the reply. thanks though
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    thank you all guys for the help i solved it!
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    (Original post by sadboynerd)
    thank you all guys for the help i solved it!
 
 
 
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