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Math Help - [Functions and Relations - logarithm

  1. #1
    Newbie jessyc's Avatar
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    Thumbs up [Functions and Relations - logarithm

    Alright, Grade 11b High school math...

    bring home assignment (fun stuff)
    Im having a lot of problems with this sheet, hopefully you guys will help me understand the problems. (assignment value is 60 counting on test marks.. due tomorrow which is Monday...)

    Id like to know if my math is correct on this question...


    3^(2x-5) = 15^(2-x)
    log 3^(2x-5) = log 15^(2-x)
    (2x-5)*log 3 = (2-x)*log 15
    2x*log 3 - 5*log 3 = 2*log 15 - x*log 15

    2x/5 = 2/x (cross multi)

    3x/3 = 10/3 = 3.3333 ....


    also i have no idea how to tackle this question...

    2/3 [12*logx a - 6(logx (4 squar root of:[b])-2*logx c^3)]

    i have more, but this is the most puzzling to me...
    Last edited by jessyc; June 3rd 2007 at 09:19 PM.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jessyc View Post
    Alright, Grade 11b High school math...

    bring home assignment (fun stuff)
    I’m having a lot of problems with this sheet, hopefully you guys will help me understand the problems. (assignment value is 60 counting on test marks.. due tomorrow which is Monday...)

    I’d like to know if my math is correct on this question...


    3^(2x-5) = 15^(2-x)
    log 3^(2x-5) = log 15^(2-x)
    (2x-5)*log 3 = (2-x)*log 15
    2x*log 3 - 5*log 3 = 2*log 15 - x*log 15

    2x/5 = 2/x (cross multi)

    3x/3 = 10/3 = 3.3333 ....
    you were good up to the third to last line, after that, i don't knw what happened. how did you even cross-multiply there?

    Picking up where you left off. i will do some rounding, so you may want to redo the steps using more decimal places if you want a more accurate answer

    2x \log 3 - 5 \log 3 = 2 \log 15 - x \log 15 ....put all the x's on one side

    \Rightarrow 2x \log 3 + x \log 15 = 5 \log 3 + 2 \log 15 ....work out the logs with your calculator

    \Rightarrow 0.9542x + 1.17609x = 2.385606 + 2.352183

    \Rightarrow 2.13029x = 4.7378

    \Rightarrow x = \frac {4.7378}{2.13029}

    \Rightarrow x = 2.224



    also i have no idea how to tackle this question...

    2/3 [12*logx a - 6(logx (4 squar root of:[b])-2*logx c^3)]
    Not sure what is going on here either.

    do you mean \frac {2}{3} \left[ 12 \log_x a - 6 \log_x \sqrt [4] {b} - 2 \log_x c^3  \right] ?

    if so, we need more info. what are a,b, and c? and what exactly do you want to find?
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  3. #3
    Newbie jessyc's Avatar
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    Sorry for
    2/3 [12*logx a - 6(logx (4 squar root of:[b])-2*logx c^3)]

    the question asks
    Express the following as a single logarithm in simplest form.

    (my math is rusty..., never good with fractions...)

    edit:
    i do have a question where values for x, y, z are given but not for logr and its asking to evaluate the expression...
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  4. #4
    Newbie jessyc's Avatar
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    Quote Originally Posted by Jhevon View Post
    ok, not a problem. but did i get the question right? once i know that we can proceed
    for 3^(2x-5) = 15^(2-x)
    yes i believe its right (i like u explaining the steps..., i wish i would write my notes like that... this will be handy when exams start in 2 weeks)
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  5. #5
    Newbie jessyc's Avatar
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    Quote Originally Posted by Jhevon View Post
    no, i mean the log question, did i type it out correctly? i didn't really explain all the steps for the first one, but if you got it, that's good


    Yes that is typed out correct.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jessyc View Post
    Yes that is typed out correct.
    ok, good

    Here are the rules you need to know for this problem.

    1. \log_a x^n = n \log_a x

    2. \log_a xy = \log_a x + \log_a y

    3. \log_a \left( \frac {x}{y} \right) = \log_a x - \log_a y



    Now, on to your problem:

    \frac {2}{3} \left[ 12 \log_x a - 6 \log_x \sqrt [4] {b} - 2 \log_x c^3  \right]


    = \frac {2}{3} \left[ \log_x a^{12} - \log_x b^{ \frac {6}{4}}  - \log_x c^6 \right] .......applied rule 1.

    = \frac {2}{3} \left[ \log_x \left( \frac {a^{12}}{b^{ \frac {3}{2}}} \right) - \log_x c^6 \right] ......applied rule 3 to the first two logs

    = \frac {2}{3} \left[ \log_x \left( \frac {a^{12}}{b^{ \frac {3}{2}} \cdot c^6} \right) \right] .......applied rule 3 including the last log

    = \log_x \left( \frac {a^{12}}{b^{ \frac {3}{2}} \cdot c^6} \right)^{ \frac {2}{3}} .......applied rule 1.

    Now you can simplify this further if you wish (you should!)

    You may be wondering, how come we never used rule 2, well, we could have but i decided not to.
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  7. #7
    Newbie jessyc's Avatar
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    Quote Originally Posted by Jhevon View Post
    ok, good

    Here are the rules you need to know for this problem.

    1. \log_a x^n = n \log_a x

    2. \log_a xy = \log_a x + \log_a y

    3. \log_a \left( \frac {x}{y} \right) = \log_a x - \log_a y



    Now, on to your problem:

    \frac {2}{3} \left[ 12 \log_x a - 6 \log_x \sqrt [4] {b} - 2 \log_x c^3  \right]


    = \frac {2}{3} \left[ \log_x a^{12} - \log_x b^{ \frac {6}{4}}  - \log_x c^6 \right] .......applied rule 1.

    = \frac {2}{3} \left[ \log_x \left( \frac {a^{12}}{b^{ \frac {3}{2}}} \right) - \log_x c^6 \right] ......applied rule 3 to the first two logs

    = \frac {2}{3} \left[ \log_x \left( \frac {a^{12}}{b^{ \frac {3}{2}} \cdot c^6} \right) \right] .......applied rule 3 including the last log

    = \log_x \left( \frac {a^{12}}{b^{ \frac {3}{2}} \cdot c^6} \right)^{ \frac {2}{3}} .......applied rule 1.

    Now you can simplify this further if you wish (you should!)

    You may be wondering, how come we never used rule 2, well, we could have but i decided not to.
    wow, , if i read/learn and understand those rules, it should help me understand almost any question like that shouldn't it?

    i read it fast, but it makes sense to me, and once again thanks for the comments on the side, they help
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jessyc View Post
    wow, , if i read/learn and understand those rules, it should help me understand almost any question like that shouldn't it?

    i read it fast, but it makes sense to me, and once again thanks for the comments on the side, they help
    yes, those are three of the four log rules that you MUST know. with those you should be able to do any log problem. When i tutor people at school i usually give them like 10 or so rules to memorize, it makes life a lot easier on them to just know those rules and apply them, but all the rules can be derived from the first four.

    in case you are wondering, the fourth law which i mentioned (which really should be the first law, since it is the definition of a logarithm) is:


    The logarithm of a number to a given base, is the power to which the base must be raised to give the number. That is,

    If \log_a b = c,

    then a^c = b
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  9. #9
    Newbie jessyc's Avatar
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    Red face great help

    Quote Originally Posted by Jhevon View Post
    yes, those are three of the four log rules that you MUST know. with those you should be able to do any log problem. When i tutor people at school i usually give them like 10 or so rules to memorize, it makes life a lot easier on them to just know those rules and apply them, but all the rules can be derived from the first four.

    in case you are wondering, the fourth law which i mentioned (which really should be the first law, since it is the definition of a logarithm) is:


    The logarithm of a number to a given base, is the power to which the base must be raised to give the number. That is,

    If \log_a b = c,

    then a^c = b
    yeah, that always helps me in class when the teacher actually goes through and shows where they derived from... although she didn't do this for logs, if she did i wasn't 'mentaly' there.

    if your up to it, i have 2 more questions, but you really really don't have to!
    (becuase where i am its really late, not sure about you)

    one i should be able to get from the laws you shown me,

    but the other is a question that is using the a(b)= x/c
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  10. #10
    Newbie jessyc's Avatar
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    this might sound kinda corny, but umm, i think you've made my night...
    and some what inspired me to keep at this 'math' thing

    edit:

    heres the new topic...
    http://www.mathhelpforum.com/math-help/urgent-homework-help/15615-b-x-b-x-getting-x-value.html#post54924

    if you're up to the challange.
    Last edited by jessyc; June 3rd 2007 at 09:18 PM. Reason: added a link to the other question
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