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Math Help - Logarithms, Algebra 2

  1. #1
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    Logarithms, Algebra 2

    You last visited: 09-17-2007 at 06:58 PM

    Sorry...the last time I visited here was when I was in like algebra 1, lol.
    I need help with some math problems about logs (I have a math test tomorrow ), I'd really appreciate the help. I'm sorry, I don't know how to make it look like the problems on my paper, but I hope this is understandable:

    log[base 5](1/8) / log[base 5](2)

    Read aloud, it reads like:

    log
    base 5
    one-eighth
    divided by
    log
    base 5
    two

    I think this would go along with log[base 5](1/8) minus log[base5](2) but I can't figure it out

    Here is another:

    7^log[base7](16) minus log[base 7](2) = x

    I was absent both the above one was done and the one below:

    log[base 7](log[base 3](log[base 2](x))) = 0

    They are both making me

    There is one last one (Sorry for the amount...I know a few of my friends can tell me how to do some in my free time before the test, but if I can get as much done here as possible, that would be awesome.)

    log (x^2) = (log x)^2

    Those logs above would be base 10


    If anyone could help I'll give you free cookies, thanks!
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  2. #2
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    Lexington, MA (USA)
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    Hello, Harryhit4!

    \frac{\log_5\left(\tfrac{1}{8}\right)} {\log_5(2)}
    There is a base-change formula: . \log_a(x) \;=\;\frac{\log_b(x)}{\log_b(a)}

    Reading it backwards, we have: . \frac{\log_5\!\left(\frac{1}{8}\right)}{\log_5(2)} \;=\;\log_2\!\left(\frac{1}{8}\right)

    Then: . \log_2\!\left(\frac{1}{8}\right) \;=\;\log_2\left(2^{-3}\right) \;=\;-3\cdot\underbrace{\log_2(2)}_{\text{This is 1}} \;=\;\boxed{-3}




    7^{\log_7(16)} - \log_7(2) \:=\:x
    This is a strange one . . . Is there a typo?

    We have: . 16 - \log_7(2) \:=\:x\quad\Rightarrow\quad \boxed{x \:=\:16-\log_7(2)}




    \log_7\bigg(\log_3\left[\log_2(x)\right]\bigg) \:=\: 0
    I hope you can follow these steps . . .

    \log_3\bigg(\log_2(x)\bigg) \:=\:7^0 \quad\Rightarrow\quad \log_3\bigg(\log_2(x)\bigg) \:=\:1

    \log_2(x) \:=\:3^1 \quad\Rightarrow\quad \log_2(x) \:=\:3

    x \:=\:2^3 \quad\Rightarrow\quad\boxed{ x \:=\:8}




    \log\left(x^2\right) \:= \:\left(\log x\right)^2

    We have: . 2\!\cdot\!\log x \:=\:\left(\log x\right)^2 \quad\Rightarrow\quad \left(\log x\right)^2 - 2\!\cdot\!\log x \:=\:0

    Factor: . \log x\,\left(\log x - 2\right) \:=\:0

    . . We have: . \log x \:=\:0 \quad\Rightarrow\quad x \:=\:10^0 \:=\:\boxed{1}

    . . . And: . \log x - 2 \:=\:0 \quad\Rightarrow\quad \log x \:=\:2\quad\Rightarrow\quad x \:=\:10^2 \:=\:\boxed{100}

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  3. #3
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    Joined
    May 2007
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    Quote Originally Posted by Soroban View Post
    Hello, Harryhit4!

    There is a base-change formula: . \log_a(x) \;=\;\frac{\log_b(x)}{\log_b(a)}

    Reading it backwards, we have: . \frac{\log_5\!\left(\frac{1}{8}\right)}{\log_5(2)} \;=\;\log_2\!\left(\frac{1}{8}\right)

    Then: . \log_2\!\left(\frac{1}{8}\right) \;=\;\log_2\left(2^{-3}\right) \;=\;-3\cdot\underbrace{\log_2(2)}_{\text{This is 1}} \;=\;\boxed{-3}




    This is a strange one . . . Is there a typo?

    We have: . 16 - \log_7(2) \:=\:x\quad\Rightarrow\quad \boxed{x \:=\:16-\log_7(2)}




    I hope you can follow these steps . . .

    \log_3\bigg(\log_2(x)\bigg) \:=\:7^0 \quad\Rightarrow\quad \log_3\bigg(\log_2(x)\bigg) \:=\:1

    \log_2(x) \:=\:3^1 \quad\Rightarrow\quad \log_2(x) \:=\:3

    x \:=\:2^3 \quad\Rightarrow\quad\boxed{ x \:=\:8}





    We have: . 2\!\cdot\!\log x \:=\:\left(\log x\right)^2 \quad\Rightarrow\quad \left(\log x\right)^2 - 2\!\cdot\!\log x \:=\:0

    Factor: . \log x\,\left(\log x - 2\right) \:=\:0

    . . We have: . \log x \:=\:0 \quad\Rightarrow\quad x \:=\:10^0 \:=\:\boxed{1}

    . . . And: . \log x - 2 \:=\:0 \quad\Rightarrow\quad \log x \:=\:2\quad\Rightarrow\quad x \:=\:10^2 \:=\:\boxed{100}

    Thank you very, very much for your help! I understand them all...

    The second one - the minus log 7 was still supposed to be in the exponent of 7. Sorry for the confusion. You probably won't be able to reply by the time I leave for school (15 mins) but I will ask my friends or my teacher. Thanks!
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