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Math Help - Logarithms

  1. #1
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    Logarithms

    Hi everyone,

    I've been given about 15 questions to work on over the holiday and I'm still in the progress of solving them. But there are a few I really cannot solve.


    1. log_3 8 = p and log_3 5 = q, express log 5 in therms of p and q
    2. If a = log_7 (11 - 6 \sqrt{2}) and b = log_7 (45 + 29 \sqrt{2}), find 3a + 2b in simplest form
    3. Solve the equation log_2 x + log_4 x + log_8 x = 11

    Thanks in advanced
    Last edited by ntrantrinh; January 5th 2010 at 06:04 PM.
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  2. #2
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    #1 I'm not sure about so I'll leave that to someone else. Some trick that is escaping me at the moment.

    #2 Remember:

    \log_{a}x^r=r\log_{a}x Apply the appropriate exponents, multiply out and use your properties of addition for logarithms to further simplify.

    #3 Is where we want to use the change of base rule:

    \log_{a}b=\frac{\log_{c}b}{\log_{c}a}

    In this instance, we want to change the log base 4 and log base 8 into something that we can merge with our log base 2. Seeing as 4 and 8 are powers of 2, this seems like an appropriate way to go:

    \log_{2}x+\frac{\log_{2}x}{\log_{2}4}+\frac{\log_{  2}x}{\log_{2}8}=11
    \log_{2}x+\frac{\log_{2}x}{2}+\frac{\log_{2}x}{3}=  11

    \frac{6\log_{2}x+3\log_{2}x+2\log_{2}x}{6}=11

    11\log_{2}x=66

    \log_{2}x=6

    2^6=x

    x=64
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  3. #3
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    Dear ntrantrinh,

    In the file attached below I have answered the first part. Please don't hesitate to reply me if you find the attachment unreadable, or if you have any further questions.

    By the way we could try the second part if you would tell us what do you mean by sqrt[29]2

    Hope this helps.
    Attached Thumbnails Attached Thumbnails Logarithms-dsc02500.jpg  
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  4. #4
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    Quote Originally Posted by Sudharaka View Post
    Dear ntrantrinh,

    In the file attached below I have answered the first part. Please don't hesitate to reply me if you find the attachment unreadable, or if you have any further questions.

    By the way we could try the second part if you would tell us what do you mean by sqrt[29]2

    Hope this helps.
    I meant \sqrt[29]{2} sorry i'll correct that now.
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  5. #5
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    Hello, ntrantrinh!

    Here's my approach to #3 . . .


    3. Solve the equation: . \log_2 x + \log_4 x + \log_8 x \:=\: 11 .[1]

    Let: . \log_2x \,=\,P \quad\Rightarrow\quad x\:=\:2^P .[2]

    Let: . \log_4x \,=\,Q \quad\Rightarrow\quad x\:=\:4^Q  \:=\:(2^2)^Q \:=\:2^{2Q} .[3]

    Let: . \log_8x \,=\,R \quad\Rightarrow\quad x \:=\:8^R \:=\:(2^3)^R \:=\:2^{3R} .[4]


    Equate [2] and [4]: . 2^P \:=\:2^{3R} \quad\Rightarrow\quad P \:=\:3R \quad\Rightarrow\quad \log_2x \:=\:3\log_8x

    Equate [3] and [4]: . 2^{2Q} \:=\:2^{3R}\quad\Rightarrow\quad 2Q \:=\:3R \quad\Rightarrow\quad Q \:=\:\tfrac{3}{2}R \quad\Rightarrow\quad \log_4x \:=\:\tfrac{3}{2}\log_8x


    Substitute into [1]: . 3\log_8x + \tfrac{3}{2}\log_8x + \log_8x \:=\:11 \quad\Rightarrow\quad \tfrac{11}{2}\log_8x \:=\:11


    Therefore: . \log_8x \:=\:2\quad\Rightarrow\quad x \:=\:8^2 \quad\Rightarrow\quad\boxed{ x \:=\:64}

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  6. #6
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    Alternatively to Soroban. \log_2(x)+\log_4(x)+\log_8(x)=\frac{\ln(x)}{\ln(2)  }+\frac{\ln(x)}{\ln\left(2^2\right)}+\frac{\ln(x)}  {\ln\left(2^3\right)} =\frac{6\ln(x)}{6\ln(2)}+\frac{3\ln(x)}{6\ln(x)}+\  frac{2\ln(x)}{6\ln(x)}=\frac{11\ln(x)}{6\ln(2)}. Therefore our equation reduces to \frac{11\ln(x)}{6\ln(2)}=11\implies \ln(x)=6\ln(2)=\ln\left(2^6\right)\implies x=2^6=64
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  7. #7
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    Hi everyone, thanks for the solutions.
    I'm still having some difficulties with the 1st and 2nd question.

    Any guidance would be greatly appreciated
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  8. #8
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    Dear ntrantrinh,

    Please clarify about what problems you are having with the first and second parts. We will try to help.
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  9. #9
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    Hello, ntrantrinh!

    Here's #2 . . .


    If . \begin{array}{ccc}a &= &\log_7 (11 - 6 \sqrt{2}) \\ b &=& \log_7 (45 + 29 \sqrt{2}) \end{array} . .find 3a + 2b in simplest form.

    We note that: . \begin{array}{ccccccc}a & = &\log(11-6\sqrt{2}) &=& (3-\sqrt{2})^2 \\<br />
b &=& \log_7(45 + 29\sqrt{2}) &=& (3+\sqrt{2})^3 \end{array}


    Then: . 3a+2b \;=\;3\log_7(3-\sqrt{2})^2 + 2\log_7(3-\sqrt{2})^3

    . . . . . . . . . . =\;6\log_7(3=\sqrt{2}) + 6\log_7(3+\sqrt{2})

    . . . . . . . . . . =\;6\bigg[\log_7(3-\sqrt{2}) + \log_7(3+\sqrt{2})\bigg]

    . . . . . . . . . . =\;6\,\log_7\!\bigg[(3-\sqrt{2})(3+\sqrt{2})\bigg]

    . . . . . . . . . . =\;6\,\underbrace{\log_7(7)}_{\text{This is 1}}

    . . . . . . . . . . =\; 6

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  10. #10
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    Quote Originally Posted by Soroban View Post
    Hello, ntrantrinh!

    Here's #2 . . .



    We note that: . \begin{array}{ccccccc}a & = &\log(11-6\sqrt{2}) &=& (3-\sqrt{2})^2 \\<br />
b &=& \log_7(45 + 29\sqrt{2}) &=& (3+\sqrt{2})^3 \end{array}


    Then: . 3a+2b \;=\;3\log_7(3-\sqrt{2})^2 + 2\log_7(3-\sqrt{2})^3

    . . . . . . . . . . =\;6\log_7(3=\sqrt{2}) + 6\log_7(3+\sqrt{2})

    . . . . . . . . . . =\;6\bigg[\log_7(3-\sqrt{2}) + \log_7(3+\sqrt{2})\bigg]

    . . . . . . . . . . =\;6\,\log_7\!\bigg[(3-\sqrt{2})(3+\sqrt{2})\bigg]

    . . . . . . . . . . =\;6\,\underbrace{\log_7(7)}_{\text{This is 1}}

    . . . . . . . . . . =\; 6

    how did you get those logs to equal (3-\sqrt{2})^2 and (3+\sqrt{2})^3
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