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Math Help - Logarithmic and exponential function questions

  1. #1
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    Exclamation Logarithmic and exponential function questions

    log3(7x+6)-log94x(squared)=1

    ey-1=6/e
    ln x2 +ey=18, find the value of (square root)ey(ln x3)
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  2. #2
    Senior Member Peritus's Avatar
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    <br />
\begin{gathered}<br />
  e^{y - 1}  = \frac{6}<br />
{e} \hfill \\<br />
   \Leftrightarrow e^{y - 1}  = e^{\ln 6 - 1}  \hfill \\<br />
   \Leftrightarrow y = \ln 6 \hfill \\ <br />
\end{gathered} <br />

    -----------------------------------------------------------
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  3. #3
    Moo
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    Hello,

    \log_3 (7x+6) = \frac{\ln(7x+6)}{\ln(3)}

    \log_9 (4x^2) = \frac{\ln(4x^2)}{\ln(9)}

    But we know that \ln(a^b) = b \ln(a)

    So, as 4x = (2x), \ln(4x)=2 \ln(2x) and as 9 = 3, \ln(9) = 2 \ln(3)

    Hence

    \log_3 (7x+6) - \log_9 (4x^2) = \frac{\ln(7x+6)}{\ln(3)} - \frac{2 \ln(2x)}{2 \ln(3)}  = \frac{\ln(7x+6)}{\ln(3)} - \frac{\ln(2x)}{\ln(3)} = \frac{\ln(7x+6) - \ln(2x)}{\ln(3)} = 1

    But ln(a)-ln(b)=ln(a/b)

    <=> \ln(\frac{7x+6}{2x}) = \ln(3)


    You'll just have to study the values of x for which the equation can't be solved.
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  4. #4
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    Hello, tyana21!

    An I reading the first one correctly?


    \log_3(7x+6) - \log_9(4x)^2 \:=\:1

    The first log can be written: . \log_3(7x+6) \;=\;2\!\cdot\!\log_9(7x+6) .**

    The second can be written: . 2\!\cdot\!\log_9(4x)


    The equation becomes: . 2\!\cdot\!\log_9(7x+6) - 2\!\cdot\!\log_9(4x) \;=\;1

    Divide by 2: . \log_9(7x+6) - \log_9(4x) \;=\;\frac{1}{2}

    Then we have: . \log_9\left(\frac{7x+6}{4x}\right) \;=\;\frac{1}{2}

    Exponentiate: . \frac{7x+6}{4x} \;=\;9^{\frac{1}{2}} \quad\Rightarrow\quad \frac{7x+6}{4x} \;=\;3

    Therefore: . 7x+6 \:=\:12x\quad\Rightarrow\quad \boxed{x \:=\:\frac{6}{5}}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    **

    Let \log_3(7x+6) \:=\:P

    Then: . 3^P \:=\:7x+6

    Take logs, base 9: . \log_9\left(3^P\right) \:=\;\log_9(7x+6) \quad\Rightarrow\quad P\!\cdot\!\log_9(3) \:=\:\log_9(7x+6)

    Since \log_9(3) = \frac{1}{2}, we have: . \frac{1}{2}P \:=\:\log_9(7x+6) \quad\Rightarrow\quad P \:=\:2\!\cdot\!\log_9(7x+6)

    Therefore: . \log_3(7x+6) \;=\;2\!\cdot\!\log_0(7x+6)

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  5. #5
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    yes you read it correctly.im a newbie so i have no idea how to write it out since i was in a rush.
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  6. #6
    Moo
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    So don't bother with my message, because i read \log_3 (7x+6) - \log_9(4x^2)=1


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  7. #7
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    Quote Originally Posted by Moo View Post
    So don't bother with my message, because i read \log_3 (7x+6) - \log_9(4x^2)=1



    haha its okay i appreciate all the help
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