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Thread: Logarithms!

  1. #1
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    Logarithms!

    $\displaystyle log_5(x+7) - log_(\sqrt5) (x+1)=1$
    Last edited by Punch; Dec 30th 2009 at 07:35 PM.
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  2. #2
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    Not sure if your latex is very helpful but

    $\displaystyle \log_5(x+7) - \log_5\frac{1}{2}(x+1) = 1$

    $\displaystyle \log_5\left(\frac{x+7}{\frac{1}{2}(x+1)}\right) = 1$

    $\displaystyle \frac{x+7}{\frac{1}{2}(x+1)} = 5$

    Can you take it from here?
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  3. #3
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    Er I think my question wasn't very clear because of LaTex...

    The 1/2 is actually the power of the second log5 which means it is $\displaystyle log_5(x+7) - log_(\sqrt5) (x+1)=1$

    The log having a base of square root is that making it difficult for me
    Last edited by Punch; Dec 30th 2009 at 07:35 PM.
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  4. #4
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    Quote Originally Posted by Punch View Post
    Er I think my question wasn't very clear because of LaTex...

    The 1/2 is actually the power of the second log5 which means it is $\displaystyle log_5(x+7) - log_(\sqrt5) (x+1)$

    The log having a base of square root is that making it difficult for me
    consider the relation between $\displaystyle log_2(x)$ and $\displaystyle log_4(y)$, how are x and y related?
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  5. #5
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    Sorry but I don'tunderstand
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  6. #6
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    $\displaystyle log_5(x+7) - log_(\sqrt5) (x+1)=1$

    but i just need to solve this equation
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  7. #7
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    Quote Originally Posted by Punch View Post
    $\displaystyle log_5(x+7) - log_(\sqrt5) (x+1)=1$

    but i just need to solve this equation
    Consider first $\displaystyle log_4(16)=2$, however if you want to take the square root of the base and preserve the same answer, you would have to take the square root of 16 as well, ergo $\displaystyle log_2(\sqrt16)=2$
    Therefore similarly you can rewrite the LHS of the original equation as:
    LHS=$\displaystyle log_5(x+7)-log_5((x+1)^2)$
    hence solve under the same base
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  8. #8
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    GAWD, So it just means that I have to square both \sqrt5 and (x+1)
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  9. #9
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    I got it, I got it yay thanks a lot!
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  10. #10
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    Hello, Punch!

    Solve for $\displaystyle x\!:\;\;\log_5(x+7) - \log_{\sqrt{5}}(x+1)\:=\:1$ .[1]

    $\displaystyle \text{Let }\log_{\sqrt{5}}(x+1) \:=\:P$ .[2]

    $\displaystyle \text{Then: }\:(\sqrt{5})^P \:=\:x+1 \quad\Rightarrow\quad \left(5^{\frac{1}{2}}\right)^P \:=\:x+1 \quad\Rightarrow\quad 5^{\frac{P}{2}} \:=\:x+1 $

    $\displaystyle \text{Take logs (base 5): }\;\log_5\left(5^{\frac{P}{2}}\right) \:=\:\log_5(x+1)$

    $\displaystyle \text{Then: }\;\frac{P}{2}\cdot\underbrace{\log_5(5)}_{\text{T his is 1}} \:=\:\log_5(x+1) \quad\Rightarrow\quad \frac{P}{2} \:=\:\log_5(x+1)$

    $\displaystyle \text{Hence: }\:P \:=\:2\log_5(x+1) \:=\:\log_5(x+1)^2$ .[3]

    $\displaystyle \text{Equate {\color{blue}[2]} and {\color{blue}[3]}: }\;\log_{\sqrt{5}}(x+1) \;=\;\log_5(x+1)^2$


    Substitute into [1]: .$\displaystyle \log_5(x+7) - \log_5(x+1)^2 \;=\;1 \quad\Rightarrow\quad \log_5\bigg[\frac{x+7}{(x+1)^2}\bigg] \;=\;1$

    Exponentiate both sides: .$\displaystyle \frac{x+7}{(x+1)^2} \:=\:5^1$

    Simplify: .$\displaystyle x+7 \:=\:5(x+1)^2 \quad\Rightarrow\quad 5x^2 + 9x - 2 \:=\:0$

    Factor: .$\displaystyle (x+2)(5x-1) \:=\:0$

    And we have two roots: .$\displaystyle x \;=\;-2,\:\tfrac{1}{5}$
    But $\displaystyle -2$ is an extraneous root.

    The solution is: .$\displaystyle \boxed{x \:=\:\frac{1}{5}}$

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  11. #11
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    thanks but what's an extraneous root?
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  12. #12
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    Smile

    Quote Originally Posted by Punch View Post
    thanks but what's an extraneous root?
    try to plug -2 in your equation and see what happens.
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  13. #13
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    lol understood thanks all for being so helpful to this maths noob here
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