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Math Help - Logarithms!

  1. #1
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    Logarithms!

    log_5(x+7) - log_(\sqrt5) (x+1)=1
    Last edited by Punch; December 30th 2009 at 08:35 PM.
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  2. #2
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    Not sure if your latex is very helpful but

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

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

     \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 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; December 30th 2009 at 08:35 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 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 log_2(x) and 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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    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
    log_5(x+7) - log_(\sqrt5) (x+1)=1

    but i just need to solve this equation
    Consider first 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 log_2(\sqrt16)=2
    Therefore similarly you can rewrite the LHS of the original equation as:
    LHS= 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 x\!:\;\;\log_5(x+7) - \log_{\sqrt{5}}(x+1)\:=\:1 .[1]

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

    \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

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

    \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)

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

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


    Substitute into [1]: . \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: . \frac{x+7}{(x+1)^2} \:=\:5^1

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

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

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

    The solution is: . \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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