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Simple logarithm question I am really stuck on: Could someone please help? watch

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    (Original post by Sam00)
    I still don't completely understand, do you mean I eliminate the power of 3/2 by squaring both sides? or square-rooting both sides?
    So by eliminating both sides you're squaring both sides like this: a^\frac{3 \times 2}{2}=8^2

    To get  a^3=64.
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    (Original post by Sam00)
    if I did (3/2)^2 i'd get 9/4 I'm confused

    You're not doing that. Are you trolling? :hmmm:

    So just looking at the indicie its  \frac{3 \times 2}{2}

    Spoiler:
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     a^\frac{3 \times 2}{2}=8^2

     a^3=64

     a=4
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    (Original post by Sam00)
    So would x^3/2 = 3√x (with 3 being the 3rd root?
    No you'd get √(a3) (square root of a3)
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    (Original post by undercxver)
    Haha, this is called \LaTeX.

    If you want help with this check out the \LaTeX help page.

    It's fun but quite long quite to do for complicated maths. :yawn:
    (Original post by Dilzo999)
    Yup \LaTeX. If you're doing maths at uni you'll probably come across it so it's good practice to use it on here .
    Much appreciated guys. Not going to do maths at uni but i've always enjoyed maths and I find it quite 'easy', so I enjoy helping others with it. I'll check it out when I can!
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    (Original post by NadeemKha_Arab)
    Much appreciated guys. Not going to do maths at uni but i've always enjoyed maths and I find it quite 'easy', so I enjoy helping others with it. I'll check it out when I can!
    Me too, I love it.

    Forgot to link the page to you!http://www.thestudentroom.co.uk/wiki/LaTex
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    (Original post by undercxver)
    Me too, I love it.

    Forgot to link the page to you!http://www.thestudentroom.co.uk/wiki/LaTex
    Yeah I had to fish it out by quoting you ahaha. I've had a look but it seems as though you have to learn all the 'functions', such as sqrt? Is there a faster way around it or is this the only way. Might take me a few tries before I get good at it
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    a^3/2 =8 is the same as sqrt a^3 = 8

    Then you can get rid of the sqrt by squaring both sides to end up with a^3 =64 (since 8^2 is 64)

    Then take the cubed root of 64 to get the value of a

    hence a =4 ( since the cube root of 64 is 4)
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    (Original post by NadeemKha_Arab)
    Yeah I had to fish it out by quoting you ahaha. I've had a look but it seems as though you have to learn all the 'functions', such as sqrt? Is there a faster way around it or is this the only way. Might take me a few tries before I get good at it
    I'm pretty sure there's a software for it, but I'm afraid you gotta use this long method on TSR as it's built in.

    Butttt, you can practice on this thread. :ahee:

    (Original post by Sam00)
    No 100% not trolling, you must think I am really thick then!

    So I can see you are squaring the 3/2 to get 3 on it's own which therefore just leaves you with a^3, and you have to also square the 8 as well which gives 64

    The thing which still confuses me is the (3x2/2) more-so the 'x2' after the 3, it doesn't seem to me you are squaring it, but more like multiplying it by 2.

    If you were squaring it wouldn't it look like 3^2/2...

    So sorry if I am not making sense
    Omg no I don't intend to call you thick! Sorry about that.

    When I said "squaring both sides" I don't mean square the power of  a once again, but I mean multiply that power.

    I think you're stuck on how indicies work. Watch these examsolutions videos.
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    (Original post by Sam00)
    No 100% not trolling, you must think I am really thick then!

    So I can see you are squaring the 3/2 to get 3 on it's own which therefore just leaves you with a^3, and you have to also square the 8 as well which gives 64

    The thing which still confuses me is the (3x2/2) more-so the 'x2' after the 3, it doesn't seem to me you are squaring it, but more like multiplying it by 2.

    If you were squaring it wouldn't it look like 3^2/2...

    So sorry if I am not making sense
    (Original post by undercxver)
    I think you're stuck on how indicies work. Watch these examsolutions videos.
    As said above, my recommendation would be to get some practice on indices. There are a few rules and once you get them stuck in your head it should be a lot easier!
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    I will try to explain here though some of the general rules based on your question:

    As we said before:
    a1/2 = √a

    Now one way to look at this is

    a3/2 = (a3)1/2

    Notice all of a3 is to the power of 1/2, meaning ALL OF a3 is square rooted.
    The reason I can take the half out is because 2 powers are multiplied together:
    e.g.

    a2 x a3 = a5

    but

    (a2)3 = a6 << (LEARN THIS RULE, THIS IS WHAT WE'RE DOING)

    So,

    (a3)1/2 = a3/2 ... only we are going in reverse to this.
 
 
 
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