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Math Help - logarithm question

  1. #1
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    logarithm question

    Hi, I need help in the following log questions:
    a) Given that log(base b)(xy^3)=m and log(base b)(x^3y^2)=p. Express log(base b)(√xy) in terms of m and p.
    b) Given 2^x=3y^2y=6^z, prove that 2xy-2yz-xz=0.

    For a. I only know adding... but cannot get log(base b) (xy).. for b. I don't really know how to start....
    Please help me in these questions, thanks.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by askh22 View Post
    a) Given that log(base b)(xy^3)=m and log(base b)(x^3y^2)=p. Express log(base b)(√xy) in terms of m and p.
    For the sake of simplicity, since we are always talking about base "b" let's just drop that symbol and use "log" to represent "log base b."

    m = log(xy^3) = log(x) + log(y^3) = log(x) + 3*log(y)
    p = log(x^3y^2) = log(x^3) + log(y^2) = 3*log(x) + 2*log(y)

    Let's change the variables a bit and call
    a = log(x)
    b = log(y)

    Then
    m = a + 3b
    p = 3a + 2b

    Now solve for a and b by your favorite method and you get:
    a = (3/7)p - (2/7)m = log(x)
    b = -(1/7)p + 3/7)m = log(y)

    We want the value of:
    log(sqrt{xy}) = (1/2)*log(xy) = (1/2)*log(x) + (1/2)*log(y)

    = (1/2)*[(3/7)p - (2/7)m] + (1/2)*[-(1/7)p + 3/7)m]

    = (1/7)p + (1/14)m

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by askh22 View Post
    2^x=3y^2y=6^z
    Is that three equations? It's too vague, I can't figure out what this could mean. Please rewrite it correctly.

    -Dan
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  4. #4
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    Appreciate so much for your prompt response & help here..

    Will get back to you the confirmed question to the question 2.. Thank for your help in advance here..
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  5. #5
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    What i had typed is same as the question paper.
    Attached Thumbnails Attached Thumbnails logarithm question-maths.png  
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  6. #6
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by askh22 View Post
    b) Given 2^x=3y^2y=6^z, prove that 2xy-2yz-xz=0.
    Solve for x:
    2^x = 3y^(2y)

    Take the ln of both sides (note: we chould just as easily take the log of both sides)
    ln2^x = ln(3y^(2y))
    xln2 = 2yln(3y)
    x = 2yln(3y)/ln2

    Solve for z:
    6^z = 3y^(2y)

    Take the ln of both sides
    ln6^z = ln(3y^(2y))
    zln6 = 2yln(3y)
    z = 2yln(3y)/ln6

    Plug x and z into 2xy-2yz-xz=0
    2[2yln(3y)/ln2]y - 2y[2yln(3y)/ln6] - [2yln(3y)/ln2]*[2yln(3y)/ln6] = 0
    4y^2*ln(3y)/ln2 - 4y^2*ln(3y)/ln6 - 4y^2*(ln(3y))^2/(ln2*ln6) = 0

    Multiply everything by the LCD = (ln2*ln6)
    4y^2*ln(3y)*ln6 - 4y^2*ln(3y)*ln2 - 4y^2*(ln(3y))^2 = 0

    Factor out 4y^2*ln(3y)
    4y^2*ln(3y)*[ln6 - ln2 - ln(3y)] = 0

    Notice that ln6 = ln(2*3) = ln2 + ln3, and that ln(3y) = ln3 + lny
    4y^2*(ln3 + lny)*(ln2 + ln3 - ln2 - ln3 - lny) = 0
    4y^2*(ln3 + lny)*(lny) = 0
    4y^2*[(lny)^2 + ln3*lny] = 0

    Using zero product property, we can set each term equal to zero:
    4y^2 = 0 --> y = 0

    (lny)^2 + ln3*lny = 0
    lny = (-ln3 +/- sqrt((ln3)^2 - 4*0))/2 = (-ln3 +/- ln3)/2 = 0 or ln3
    y = e^0 or e^(ln3) --> y = 1 or 3

    So, y = 0, 1, 3

    Solving for the value of x, we get:
    y = 0: x = 2(0)ln(3*0)/ln2 = undefined
    y = 1: x = 2(1)ln(3)/ln2 = ln9/ln2
    y = 3: x = 2(3)ln(9)/ln2 = 6ln9/ln2

    Solving for the value of z, we get:
    y = 0: z = 2(0)ln(3*0)/ln6 = undefined
    y = 1: z = 2(1)ln(3)/ln6 = ln9/ln6
    y = 3: z = 2(3)ln(9)/ln6 = 6ln9/ln6

    I'm not sure this "proves" anything , except that there are at least 2 points where the above is true.

    (I was tempted to delete everything I just did but I figured it might be helpful.)
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by askh22 View Post
    What i had typed is same as the question paper.
    As long as there are parenthesis! It was the 3y^2y term that was causing my confusion.

    This probably won't be the best method, but it works.

    So:
    2^x=3*y^{2y}=6^z

    We want to show that
    2xy-2yz-xz=0
    is an identity.

    Let's consider
    2^x = 6^z

    Take the natural log (or log base 10 or any other base) of both sides:
    ln(2^x) = ln(6^z)

    x*ln(2) = z*ln(6)

    So
    z = [ln(2)/ln(6)]*x

    Thus
    2xy - 2yz - xz = 0
    implies
    2xy - 2y[ln(2)/ln(6)]*x - x[ln(2)/ln(6)]*x = 0

    We can now factor an x:
    x*(2y - 2y[ln(2)/ln(6)] - x[ln(2)/ln(6)]) = 0

    If x is 0 then this is true. So one solution is x = 0, which implies z = 0, which implies:
    3*y^{2y} = 1

    y^{2y} = 1/3 <-- Take the square root of both sides.

    y^y - sqrt(1/3) = 0

    We can graphically see that this function has no zeros. Thus x cannot be zero.

    If we assume x is not 0 then
    2y - 2y[ln(2)/ln(6)] - x[ln(2)/ln(6)] = 0

    2^x=3*y^{2y}

    Again take the natural log of both sides:
    x*ln(2) = ln[3*y^{2y}] = ln(3) + ln[y^{2y}] = ln(3) + (2y)*ln(y)

    So
    x = ln(3)/ln(2) + (2/ln(2))*y*ln(y)

    Inserting this into the top equation:
    2y - 2y[ln(2)/ln(6)] - [ln(3)/ln(2) + (2/ln(2))*y*ln(y)]*[ln(2)/ln(6)] = 0

    2y - 2y[ln(2)/ln(6)] - ln(3)/ln(6) - (2*ln(2)/ln(6))*y*ln(y) = 0

    The problem is that I only know how to solve this numerically. I get a zero around y = 0.2759

    So
    x = ln(3)/ln(2) + (2/ln(2))*y*ln(y) --> x = 0.5598

    z = [ln(2)/ln(6)]*x --> z = 0.2166

    So finally I get
    x = 0.5598
    y = 0.2759
    z = 0.2166

    Notes:
    1) I never verified that there is only one solution for y. You should check this.

    2) You need a few more digits in the approximation to get good accuracy on 2xy-2yz-xz=0.

    -Dan
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ecMathGeek View Post
    Take the ln of both sides (note: we chould just as easily take the log of both sides)
    ln2^x = ln(3y^(2y))
    xln2 = 2yln(3y)
    You did the same thing I did the first time around. The "3" is not raised to the (2y) power in the expression, so
    xln2 = ln(3*y^{2y}) = ln(3) + ln(y^{2y}) = ln(3) + (2y)*ln(y)

    -Dan
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  9. #9
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by topsquark View Post
    You did the same thing I did the first time around. The "3" is not raised to the (2y) power in the expression, so
    xln2 = ln(3*y^{2y}) = ln(3) + ln(y^{2y}) = ln(3) + (2y)*ln(y)

    -Dan
    Thank you. I didn't see that I was doing that. No wonder it was so hard .
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  10. #10
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    Hi Dan & ecMathGeek,
    Thanks for the reply, maybe the question did not explain clearly. What it asking is to show that with 2^x=3y^2y=6^z, we can derive 2xy-2yz-xz=0, no solving is needed. That's why i getting stuck. The solutions you guys provided are too hard for me
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  11. #11
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by askh22 View Post
    Hi Dan & ecMathGeek,
    Thanks for the reply, maybe the question did not explain clearly. What it asking is to show that with 2^x=3y^2y=6^z, we can derive 2xy-2yz-xz=0, no solving is needed. That's why i getting stuck. The solutions you guys provided are too hard for me
    The problem is, according to what topsquark has shown (and what I attempted to show, outside of the mistakes I made), there is (at least) one specific answer to this problem - one case where it is true, which suggests that it may not be true for all values of x, y, and z.

    2^x = 2y^(2y) = 6^z does not necessarily result in 2xy - 2yz - xz = 0, except for with specific values of x, y, and z.
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  12. #12
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    Quote Originally Posted by ecMathGeek View Post
    The problem is, according to what topsquark has shown (and what I attempted to show, outside of the mistakes I made), there is (at least) one specific answer to this problem - one case where it is true, which suggests that it may not be true for all values of x, y, and z.

    2^x = 2y^(2y) = 6^z does not necessarily result in 2xy - 2yz - xz = 0, except for with specific values of x, y, and z.
    I agree. However this type of question that I had encounter has already unofficial specified the statement 2xy - 2yz - xz = 0 is true, and it is asking what is the steps to achieve it with the condition 2^x = 2y^(2y) = 6^z.
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  13. #13
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by askh22 View Post
    I agree. However this type of question that I had encounter has already unofficial specified the statement 2xy - 2yz - xz = 0 is true, and it is asking what is the steps to achieve it with the condition 2^x = 3y^(2y) = 6^z.
    The problem is that it is NOT automatic. We would have to be able to come up with specific values of x, y, and z only from the relations 2^x = 3y^(2y) = 6^z and then show that those values satisfy 2xy - 2yz - xz = 0.

    Let me show you what happens. We have
    2^x = 3y^(2y)
    2^x = 6^z
    3y^(2y) = 6^z

    But the third relation is implied by the first two. So we only have two independent equations in three variables. So the best we can do is to solve for x and z as functions of y. Let's do that:
    2^x = 3y^(2y)

    ln[2^x] = ln[3y^{2y}]

    x*ln(2) = ln(3) + 2y*ln(y)

    x = ln(3)/ln(2) + (2/ln(2))*y*ln(y)

    We also have
    6^z = 3y^(2y)

    z = ln(3)/ln(6) + (2/ln(6))*y*ln(y)

    From this you would need to show that for all y (in the domian):
    2xy - 2yz - xz = 0

    2*[ln(3)/ln(2) + (2/ln(2))*y*ln(y)] - 2*y*[ln(3)/ln(6) + (2/ln(6))*y*ln(y)] - [ln(3)/ln(2) + (2/ln(2))*y*ln(y)]*[ln(3)/ln(6) + (2/ln(6))*y*ln(y)] = 0

    This simplifies a bit, but we can already show that disaster looms. This expression should be true for all real y (not equal to 0). Set y = 1:
    2*[ln(3)/ln(2) + (2/ln(2))*ln(1)] - 2*[ln(3)/ln(6) + (2/ln(6))*ln(1)] - [ln(3)/ln(2) + (2/ln(2))*ln(1)]*[ln(3)/ln(6) + (2/ln(6))*ln(1)] = 0

    Now, ln(1) = 0 so....
    2*[ln(3)/ln(2)] - 2*[ln(3)/ln(6)] - [ln(3)/ln(2)]*[ln(3)/ln(6)] = 0

    0.971815 = 0

    which is obviously not true.

    So we can find a set of (x, y, z) such that 2xy - 2yz - xz = 0 given the conditions 2^x = 3y^(2y) = 6^z, but it will not be true in general.

    -Dan
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  14. #14
    Senior Member ecMathGeek's Avatar
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    Either way we look at this problem, we cannot help but be restricted by the fact that 2^x = 3y^(2y) = 6^z constricts the domain on which particular solutions to 2xy - 2yz - xz = 0 exist. The one relation cannot be said to equal the other relation except for at specific (and most likely finite) points (x,y,z).

    The relationship 2^x = 3y^(2y) = 6^z represents a "curve" in 3D space, and 2xy - 2yz - xz = 0 represents a "surface" in 3D space. We can find solutions to this only where the curve 2^x = 3y^(2y) = 6^z intersects the surface 2xy - 2yz - xz = 0. I suspect this occurs at most twice over the domain of each function. And so all we can do with this problem (that I know of) is solve it to find the particular "solutions," one of which topsquark has found, that satisfy these two relationships.

    In other words, it cannot be proven that given this relationship, 2^x = 3y^(2y) = 6^z, this, 2xy - 2yz - xz = 0, MUST be true.
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