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Math Help - log question (last one)

  1. #1
    Newbie jessyc's Avatar
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    Post log question (last one)

    thought i could do it... need sleep thats all

    given that log r X = -5, log r Y = 2 and log r Z = 9, evaluate the expression

    log r ( r^4 x-1 y^5 / 3[sqroot: x] )
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  2. #2
    Newbie jessyc's Avatar
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    Quote Originally Posted by Jhevon View Post
    i assume you mean log to the base r when you write log r

    what is that x - 1? should it be x^-1? please use parenthesis to make your question clear

    logr ( {r^4} {x^-1} {y^5} / 3[sqroot: x] )

    edit, base of r , you where correct
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jessyc View Post
    given that log r X = -5, log r Y = 2 and log r Z = 9, evaluate the expression

    logr ( {r^4} {x^-1} {y^5} / 3[sqroot: x] )
    i think you meant cube root of x

    \log_r \left( \frac {r^4 x^{-1} y^5}{ \sqrt [3] {x}} \right)

    = \log_r \left( \frac {r^4 y^5}{x \sqrt [3] {x}} \right) .......since x^{-1} = \frac {1}{x}

    = \log_r \left( r^4 y^5 \right) - \log_r \left( x^{ \frac {4}{3}} \right) ......since \log_a \left( \frac {x}{y} \right) = \log_a x - \log_a y

    = \log_r r^4 + \log_r y^5 - \log_r x^{ \frac {4}{3}} .....since \log_a xy = \log_a x + \log_a y

    = 4 \log_r r  + 5 \log_r y - \frac {4}{3} \log_r x .......since \log_a x^n = n \log_a x

    = 4(1) + 5(2) - \frac {4}{3}(-5) ...........since \log_a a = 1

    = 4 + 10 + \frac {20}{3}

    I leave the last step to you


    EDIT: If you have any questions, say so. I subtly applied some of the laws of exponents, make sure you don't miss them
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  4. #4
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    Quote Originally Posted by jessyc View Post
    logr ( {r^4} {x^-1} {y^5} / 3[sqroot: x] )

    edit, base of r , you where correct
    Hello,

    I assume that there is a typo and that you mean:

    \log_r \left(\frac{z^4 \cdot x^{-1} \cdot y^5}{3 \cdot \sqrt{x}} \right) = 4 \cdot \log_r(z) - \log_r(x) + 5 \cdot \log_r(y) - \frac{3}{2} \cdot \log_r(x) Now plug in the values you know:

    = 4 \cdot 9 - (-5) + 5 \cdot 2 - \frac{3}{2} \cdot (-5) = 58.5
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by earboth View Post
    Hello,

    I assume that there is a typo and that you mean:

    \log_r \left(\frac{z^4 \cdot x^{-1} \cdot y^5}{3 \cdot \sqrt{x}} \right) = 4 \cdot \log_r(z) - \log_r(x) + 5 \cdot \log_r(y) - \frac{3}{2} \cdot \log_r(x) Now plug in the values you know:

    = 4 \cdot 9 - (-5) + 5 \cdot 2 - \frac{3}{2} \cdot (-5) = 58.5
    yeah, i was wondering about that r being there and no z

    i don't think it would be 3 \cdot \sqrt {x} though, since you would end up with log_r 3 + \frac {1}{2} \log_r x \neq \frac {3}{2} \log_r x
    Last edited by Jhevon; June 4th 2007 at 08:54 AM.
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  6. #6
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    Hello, jessyc!

    More questions . . .


    Given that: .log r X = -5, log r Y = 2 and log r Z = 9, evaluate:

    log r ( r^4 x^-1 y^5 / 3[sqroot: x] )

    I assume we are given: . \log_r(x) = -5,\;\log_r(Y) = 2,\;\log_r(Z) = 9

    . . and: . \log_r\left(\frac{r^4\!\cdot\!X^{-1}\!\cdot\!Y^5}{3\sqrt{X}}\right)


    But I bet the denominator is: \sqrt[3]{Z}

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  7. #7
    Newbie jessyc's Avatar
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    this is my work so far i'm rewriting the questions so they look better
    this is proof of my work...

    the pages are double sided... anyways thanks

    edit: oh and its 2:45 AM got school at 9 have to get up at 8 love life (thanks for making it easyer guys )
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  8. #8
    Newbie jessyc's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, jessyc!

    More questions . . .



    I assume we are given: . \log_r(x) = -5,\;\log_r(Y) = 2,\;\log_r(Z) = 9

    . . and: . \log_r\left(\frac{r^4\!\cdot\!X^{-1}\!\cdot\!Y^5}{3\sqrt{X}}\right)


    But I bet the denominator is: \sqrt[3]{Z}

    how would i do it?
    ...
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  9. #9
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jessyc View Post
    how would i do it?
    ...
    exactly the way you saw us do the last one. except at the second to last line, you would plug in the value for Z instead. you should try it and tell us your answer
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  10. #10
    Newbie jessyc's Avatar
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    it was late last night....

    logr ({r^4}{y^5}/x{3rd root of z}) = 1/x

    logr^4{y^5} = x * {3rd root of z}

    logr^4 = x * {3rd root of z} / y^5

    r^4 = 0.3250
    (then just rais to the 4th power.)

    this answer doesn't look right to me...
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  11. #11
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    Quote Originally Posted by jessyc View Post
    it was late last night....

    logr ({r^4}{y^5}/x{3rd root of z}) = 1/x

    logr^4{y^5} = x * {3rd root of z}

    logr^4 = x * {3rd root of z} / y^5

    r^4 = 0.3250
    (then just rais to the 4th power.)

    this answer doesn't look right to me...
    Hello, jessyc,

    I read your problem like this:

    \log_r \left(\frac{r^4 \cdot y^5}{x \cdot \sqrt[3]{z}}  \right)= \frac{1}{x}

    I'm not quite sure what you want to do with this equation... So I'll try to simplify the LHS of it:

    4 + 5\log_r(y)-\log_r(x) - \frac{1}{3} \cdot \log_r(z) = \frac{1}{x}

    Could it be that the RHS of this equation reads \log_r \left(\frac{1}{x} \right) = -\log_r(x) ?
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  12. #12
    Newbie jessyc's Avatar
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    sorry x,y,z have values,
    im not sure what i was suppose to do with it either.
    log r X = -5, log r Y = 2 and log r Z = 9,
    are the values

    i was assuming i was trying to find r...
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  13. #13
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jessyc View Post
    sorry x,y,z have values,
    im not sure what i was suppose to do with it either.
    log r X = -5, log r Y = 2 and log r Z = 9,
    are the values

    i was assuming i was trying to find r...
    i think finding r exactly is pretty much impossible here, there are too many unknowns. as earboth suggested, they probably want us to prove that the left hand side is the same as the right hand side...or wait! maybe not! maybe we could find r....

    pick up where earboth left off.

    4 + 5\log_r(y)-\log_r(x) - \frac{1}{3} \cdot \log_r(z) = \frac{1}{x}

    \Rightarrow 4 + 5 (2) - (-5) - \frac {1}{3} 9 = \frac {1}{x}

    \Rightarrow 16 = \frac {1}{x}

    \Rightarrow x = \frac {1}{16}

    Now we are told that \log_r x = -5

    \Rightarrow \log_r \left( \frac {1}{16} \right) = -5

    \Rightarrow r^{-5} = \frac {1}{16} .......since, if \log_a b = c \Rightarrow a^c = b

    \Rightarrow r^5 = 16

    \Rightarrow r = \sqrt [5] {16}

    and you can plug that into your calculator to see what it is
    Last edited by Jhevon; June 4th 2007 at 09:26 AM. Reason: Silly mistake, thanks Soroban
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  14. #14
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    Hello, jessyc!

    I assume I've incorporated the corrections . . .


    Given: . \log_r(X) = -5,\;\;\log_r(Y) = 2,\;\;\log_r(Z) = 9

    . . Evaluate: . \log_r\!\left(\frac{r^4\!\cdot\!X^{-1}\!\cdot\!Y^5}{\sqrt[3]{Z}}\right)

    We have: . \log_r\!\left(r^4\!\cdot\!X^{-1}\!\cdot\!Y^5\right) - \log_r\!\left(\sqrt[3]{Z}\right)

    . . = \;\log_r\!\left(r^4\right) + \log_r\!\left(X^{-1}\right) + \log_r\!\left(Y^5\right) - \log_r\!\left(Z^{\frac{1}{3}}\right)

    . . = \;4\!\cdot\!\underbrace{\log_r(r)}_{\downarrow} \,-\,1\!\cdot\!\underbrace{\log_r(X)}_{\downarrow} \,+\, 5\!\cdot\!\underbrace{\log_r(Y)}_{\downarrow} \,-\, \frac{1}{3}\!\cdot\!\underbrace{\log_r(Z)}_{\downa  rrow}
    . . = \;\;\;4\cdot(1) \;\;-\;\; 1\cdot(\text{-}5) \;\;\;+\;\;\; 5\cdot(2) \;\;\;\,-\,\;\;\; \frac{1}{3}\cdot(9)

    Answer: . 4 + 5 + 10 - 3 \;=\;\boxed{16}

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