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Math Help - Latex Help Needed

  1. #1
    Member cmf0106's Avatar
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    Latex Help Needed


    Hey Guys I need help turning this into LATEX, I have tried at it for a good portion of the night but cant seem to find the correct combination.

    Here is what I have so far if anyone would like to correct it.
    10.(\frac{(x+8)^2}{x^4})^{-3}=\frac{(x^4)^3}{[(x+8)^2]^3}=
    \frac{x^{12}{(x+8)^6}}
    11.(\frac{(x+3)^{-2}}{y^{-4}})^{-3}=\frac{[(x+3)^{-2}]^{-3}}{(y^{-4}0^{-3}}=\frac{(x+3)}^{(-2)}{(-3)}{y^{(-4)}{(-3)}}=\frac{(x+3)^6}{y^{12}}

    *with [tex] tags removed of course*


    Many thanks
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  2. #2
    o_O
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    #10:

    This is what you have so far (adjusted your coding for a bit):
    (\frac{(x+8)^2}{x^4})^{-3}=\frac{(x^4)^3}{[(x+8)^2]^3}=<br />
\frac{x^{12}}{(x+8)^6}

    from: (\frac{(x+8)^2}{x^4})^{-3}=\frac{(x^4)^3}{[(x+8)^2]^3}=
    \frac{x^{12}}{(x+8)^6} (removed some extra braces you had here and there as well)

    Looks good to me except for that first parantheses. What you can do is use \left( ... \right) in place of the regular ( ... ) to get:
    \left(\frac{(x+8)^2}{x^4}\right)^{-3}=\frac{(x^4)^3}{[(x+8)^2]^3}=<br />
\frac{x^{12}}{(x+8)^6}

    This applies to brackets as well: \left[ ... \right], \left| ... \right|, etc. etc.
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  3. #3
    o_O
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    #11 is a doosie. There are some random typos that you should look over but it seems the main problem is in how you're dealing with your exponents.

    To get put an exponent over a character, you simply but ^{ stuff } right after the character. For example, y^{(-3)(-2)} is generated from y^{(-3)(-2)} (bolded for emphasis).

    What you did y^{(-2)}{(-3)} will not give you the desired exponent because you ended it by the first right brace. This will give you: y^{(-2)}{(-3)}

    Try it again and see what you can come up with and post back if you have problems.
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  4. #4
    Member cmf0106's Avatar
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    I'll give 11 a shot perhaps in the morning, I am too mentally exhausted to attempt it now. I really appreciate your help though but if any one else if willing to give 11 a shot, please by all means do so.

    Alternatively I can just leave a blank space on my review sheet where #11 is supposed to go and hand write it in.
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by cmf0106 View Post

    Hey Guys I need help turning this into LATEX, I have tried at it for a good portion of the night but cant seem to find the correct combination.

    Here is what I have so far if anyone would like to correct it.
    10.(\frac{(x+8)^2}{x^4})^{-3}=\frac{(x^4)^3}{[(x+8)^2]^3}=
    \frac{x^{12}}{(x+8)^6}
    11.(\frac{(x+3)^{-2}}{y^{-4}})^{-3}=\frac{[(x+3)^{-2}]^{-3}}{(y^{-4}0^{-3}}=\frac{(x+3)}^{(-2)}{(-3)}{y^{(-4)}{(-3)}}=\frac{(x+3)^6}{y^{12}}

    *with [tex] tags removed of course*


    Many thanks
    You left out the part in red. When in the math tags, you would have had an error. Also, you had an extra } in the code somewhere...

    You can use \bigg to make the parenthesis\brackets larger

    so you could have

    Code:
    [tex]\bigg(\frac{(x+8)^2}{x^4}\bigg)^{-3}=\frac{(x^4)^3}{[(x+8)^2]^3}=
    \frac{x^{12}}{(x+8)^6}[/tex]
    which gives

    \bigg(\frac{(x+8)^2}{x^4}\bigg)^{-3}=\frac{(x^4)^3}{[(x+8)^2]^3}=<br />
\frac{x^{12}}{(x+8)^6}

    You can make some of the same adjustments for the other one.

    However I see some faults in the code for the second one.

    (\frac{(x+3)^{-2}}{y^{-4}})^{-3}=\frac{[(x+3)^{-2}]^{-3}}{(y^{-4}0^{-3}}=\frac{(x+3)}^{(-2)}{(-3)}{y^{(-4)}{(-3)}}=\frac{(x+3)^6}{y^{12}}

    That should be a parenthesis: )

    There are a couple things that would give you a syntax error:

    (\frac{(x+3)^{-2}}{y^{-4}})^{-3}=\frac{[(x+3)^{-2}]^{-3}}{(y^{-4}0^{-3}}=\frac{(x+3)}^{(-2)}{(-3)}{y^{(-4)}{(-3)}}=\frac{(x+3)^6}{y^{12}}

    To get them in one exponent, just leave it as {(-2)(-3)} and {(-4)(-3)}.

    The code should be [after making the larger parenthesis adjustment]:

    \bigg(\frac{(x+3)^{-2}}{y^{-4}}\bigg)^{-3}=\frac{[(x+3)^{-2}]^{-3}}{(y^{-4})^{-3}}=\frac{(x+3)}^{(-2)(-3)}{y^{(-4)(-3)}}=\frac{(x+3)^6}{y^{12}}

    However, there is still an error:

    \bigg(\frac{(x+3)^{-2}}{y^{-4}}\bigg)^{-3}=\frac{[(x+3)^{-2}]^{-3}}{(y^{-4})^{-3}}=\frac{(x+3)}^{(-2)(-3)}{y^{(-4)}{(-3)}}=\frac{(x+3)^6}{y^{12}}

    That brace is in the wrong place:

    \bigg(\frac{(x+3)^{-2}}{y^{-4}}\bigg)^{-3}=\frac{[(x+3)^{-2}]^{-3}}{(y^{-4})^{-3}}=\frac{(x+3)}^{(-2)(-3)}}{y^{(-4)}{(-3)}}=\frac{(x+3)^6}{y^{12}}

    The brace should be in the place where the blue one is.

    So your code should now be:

    \bigg(\frac{(x+3)^{-2}}{y^{-4}}\bigg)^{-3}=\frac{[(x+3)^{-2}]^{-3}}{(y^{-4})^{-3}}=\frac{(x+3)^{(-2)(-3)}}{y^{(-4)(-3)}}=\frac{(x+3)^6}{y^{12}}

    which yields:

    \bigg(\frac{(x+3)^{-2}}{y^{-4}}\bigg)^{-3}=\frac{[(x+3)^{-2}]^{-3}}{(y^{-4})^{-3}}=\frac{(x+3)^{(-2)(-3)}}{y^{(-4)(-3)}}=\frac{(x+3)^6}{y^{12}}

    also, you can use \left( ... \right) for the parenthesis! Keep that in mind.

    I hope this helps!

    --Chris

    EDIT: I see that o_O beat me again...I must be a slow typer
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