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    solve each equation givingyour answers in the form 'ln r' where 'r' is a rational number

    b) 5 + 4e^2x = 6

    my working out:

    5 + 4e^2x = 6
    4e^2x = 1
    e^2x=1/4
    lne^2x = ln1/4
    2x = ln1/4
    x = -0.693....

    the answer is = ln1/2

    any help would be great!
    thankkks
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    (Original post by boomboompow.)
    solve each equation givingyour answers in the form 'ln r' where 'r' is a rational number

    b) 5 + 4e^2x = 6

    my working out:

    5 + 4e^2x = 6
    4e^2x = 1
    e^2x=1/4
    lne^2x = ln1/4
    2x = ln1/4
    x = -0.693....

    the answer is = ln1/2

    any help would be great!
    thankkks
    ln1/4 divided by 2 is ln 1/2
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    (Original post by boomboompow.)
    ...2x = ln1/4...
    Therefore x=\frac{1}{2}\ln (\frac{1}{4})
    Now use the law of logs which states that:
    a\ln (b) = \ln (b^a).
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    (Original post by Doughnuts!!)
    ln1/4 divided by 2 is ln 1/2
    i'm such an idiot...... hah
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    (Original post by boomboompow.)
    solve each equation givingyour answers in the form 'ln r' where 'r' is a rational number

    b) 5 + 4e^2x = 6

    my working out:

    5 + 4e^2x = 6
    4e^2x = 1
    e^2x=1/4
    lne^2x = ln1/4
    2x = ln1/4
    x = -0.693....

    the answer is = ln1/2

    any help would be great!
    thankkks
    ln(1/2) is -0.693...

    I know this off by heart for some reason. :|

    :indiff:
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    (Original post by Farhan.Hanif93)
    Therefore x=\frac{1}{2}\ln (\frac{1}{4})
    Now use the law of logs which states that:
    a\ln (b) = \ln (b^a).
    Hey dude! :hello:

    I was just gonna say that surely it would just be easier to say that x=\frac{1}{2}\ln (\frac{1}{4}) is the same as saying ln1/4 divided by 2?
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    (Original post by Doughnuts!!)
    Hey dude! :hello:

    I was just gonna say that surely it would just be easier to say that x=\frac{1}{2}\ln (\frac{1}{4}) is the same as saying ln1/4 divided by 2?
    Hey.
    Not sure what you mean by that? It's not obvious if you don't know your laws of logs.
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    (Original post by Farhan.Hanif93)
    Hey.
    Not sure what you mean by that? It's not obvious if you don't know your laws of logs.
    As in 1/2(ln 1/4) is exactly the same as ln1/4 divided by 2? Is it not? I think I'm wrong now and I can see why you used log laws to do it...

    Sorry for not using latex btw! :p:
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    (Original post by Doughnuts!!)
    As in 1/2(ln 1/4) is exactly the same as ln1/4 divided by 2? Is it not? I think I'm wrong now and I can see why you used log laws to do it...

    Sorry for not using latex btw! :p:
    Are you trying to say " ln(1/4) divided by 2 "

    or

    " ln ( (1/4) divded by 2) "
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    (Original post by Doughnuts!!)
    As in 1/2(ln 1/4) is exactly the same as ln1/4 divided by 2? Is it not? I think I'm wrong now and I can see why you used log laws to do it...

    Sorry for not using latex btw! :p:
    That's true. But I'm not sure why it's relevant here?
    The only way to show it is by using that law I mentioned above.
    EDIT: If it's what ziedj has said, then it's just a coincidence that it is true in this case! :p: Well spotted though.
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    (Original post by ziedj)
    Are you trying to say " ln(1/4) divided by 2 "

    or

    " ln ( (1/4) divded by 2) "
    ln(1/4) divided by 2
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    (Original post by Farhan.Hanif93)
    That's true. But I'm not sure why it's relevant here?
    The only way to show it is by using that law I mentioned above.
    EDIT: If it's what ziedj has said, then it's just a coincidence that it is true in this case! :p: Well spotted though.
    Well considering (1/4) divided by 2 = 1/8, I'm not so sure!
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    (Original post by Farhan.Hanif93)
    That's true. But I'm not sure why it's relevant here?
    The only way to show it is by using that law I mentioned above.
    EDIT: If it's what ziedj has said, then it's just a coincidence that it is true in this case! :p: Well spotted though.
    Ahhh, makes sense now. It won't work in all cases like this, as it can only be proved by log laws! This was just a coincidence!

    That's cleared things up!
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    (Original post by ziedj)
    Well considering (1/4) divided by 2 = 1/8, I'm not so sure!
    I don't know what he's spotted but if he divided by half then the answer comes out correctly. I'm just making things up now to fit though haha!
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    (Original post by Doughnuts!!)
    ln(1/4) divided by 2
    Okay, so instead of putting the fraction at the front you put it underneath.. how does that help you get the answer?

    What you should be doing is noting that \frac{1}{2} \log \frac{1}{4} = \log \frac{1}{4}^\frac{1}{2} = \log \sqrt{ \frac{1}{4}} = \log \frac{1}{2}
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    (Original post by Farhan.Hanif93)
    I don't know what he's spotted but if he divided by half then the answer comes out correctly. I'm just making things up now to fit though haha!
    Haha, yeah making things up to fit was what I was doing all along!

    Yeah, it works if you divide it by 1/2, rather than 2. :yep:
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    (Original post by ziedj)
    Okay, so instead of putting the fraction at the front you put it underneath.. how does that help you get the answer?

    What you should be doing is noting that \frac{1}{2} \log \frac{1}{4} = \log \frac{1}{4}^\frac{1}{2} = \log \sqrt{ \frac{1}{4}} = \log \frac{1}{2}
    Yeah, I've realised that now! :o:
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    (Original post by -beads-)
    ln(1/2) is -0.693...

    I know this off by heart for some reason. :|

    :indiff:
    that's cos you did physics right? :lol:

    OP i got ln(0.5)=-0.693
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    (Original post by Doughnuts!!)
    Yeah, I've realised that now! :o:
    At least you realised, and now know the correct method - loads of people would happily carry on assuming you can just "divide the inside bit by the fraction at the front and it will always work.. I saw it work once!"

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    (Original post by Remarqable M)
    that's cos you did physics right? :lol:

    OP i got ln(0.5)=-0.693
    I do, but you don't use ln(x) in Physics?

    It's just that i've done my A-level in Maths and I used ln(0.5) a lot. :laugh:
 
 
 
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