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Thread: Exponential and Logarithmic Functions help!!!

  1. #1
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    Exponential and Logarithmic Functions help!!!

    Hi everyone!

    I am currently having trouble understanding a couple parts of my homework. I filled in the rest but these got confusing:

    C. Change each equation to its exponential form:

    $\displaystyle lne^2$

    D. Change each equation to its Log form. Assume y > 0 and b > 0.

    $\displaystyle y = e^x$

    Answer:
    Quote Originally Posted by ThePerfectHacker View Post
    If $\displaystyle y=e^x$ then $\displaystyle x = \ln y$.
    E. Evaluate each log.

    $\displaystyle lne^3$

    Answer:
    Quote Originally Posted by ThePerfectHacker View Post
    $\displaystyle \ln e^n = n$ because $\displaystyle e^{(n)} = e^n$.
    Now this is the part I am more confused about

    F. Solve each exponential equation.

    $\displaystyle 0.3(4^0.2x) = 0.2$

    $\displaystyle 9^x - 3^(x+1) + 1 = 0$

    $\displaystyle log2(3x + 2) - log4x = 3$

    $\displaystyle (e^x + e^-x) / 2 = 3$


    You can answer them for me but please explain to me how you got the answer or what I need to do to get the answer.

    I'm still working on these problems and finished a few of them but I want to see if I am positively correct.

    Thank you guys!

    --Alex
    Last edited by RedSpades; Oct 24th 2007 at 05:55 PM. Reason: Better editing and added answers...
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  2. #2
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    Quote Originally Posted by RedSpades View Post
    Hi everyone!

    I am currently having trouble understanding a couple parts of my homework. I filled in the rest but these got confusing:

    C. Change each equation to its exponential form:

    ln e^2

    D. Change each equation to its Log form. Assume y > 0 and b > 0.

    y = e^x

    E. Evaluate each log.

    ln e^3

    Now this is the part I am more confused about

    F. Solve each exponential equation.

    0.3(4^0.2x) = 0.2

    9^x - 3^(x+1) + 1 = 0

    log2(3x + 2) - log4x = 3

    (e^x + e^-x) / 2 = 3


    You can answer them for me but please explain to me how you got the answer or what I need to do to get the answer.

    I'm still working on these problems and finished a few of them but I want to see if I am positively correct.

    Thank you guys!

    --Alex
    If $\displaystyle y=e^x$ then $\displaystyle x = \ln y$.

    And $\displaystyle \ln e^n = n$ because $\displaystyle e^{(n)} = e^n$.
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  3. #3
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    Thank you PerfectHacker for the answers.

    I looked at the answers and I figured how it worked. Thanks!

    Anyone else know how to work with the other problems, especially section F?

    edit: I edited the main post and added the answers and made the problems appear more clearer. :P
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  4. #4
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    Hello, Alex!

    Could you re-type part F?


    F. Solve each exponential equation.

    $\displaystyle [1] \;\;0.3(4^0.2x) = 0.2$
    Can't read it . . .

    Is that: .$\displaystyle 0.3(4^{0.2x}) \:=\:0.2 $ ?



    $\displaystyle [2]\;\;9^x - 3^(x+1) + 1 = 0$
    I assume that it is: .$\displaystyle 9^x - 3^{x+1} + 1 \:=\:0$

    You need to enclose exponents in brackets: .3^{x+1}

    We have: .$\displaystyle \left(3^2\right)^x - 3^x\!\cdot\!3 + 1 \:=\:0$

    . . then: .$\displaystyle \left(3^x\right)^2 - 3\left(3^x\right) + 1\:=\:0$

    Let $\displaystyle u \,=\,3^x$ and we have: .$\displaystyle u^3 - 3u + 1 \:=\:0$

    Quadratic Formula: .$\displaystyle u \;=\;\frac{3\pm\sqrt{3^2-4(1)(1)}}{2(1)} \;=\;\frac{3\pm\sqrt{5}}{2}$

    Back-substitute: .$\displaystyle 3^x \:=\:\frac{3\pm\sqrt{5}}{2}$

    Therefore: .$\displaystyle x \;=\;\log_3\left(\frac{3\pm\sqrt{5}}{2}\right)$



    $\displaystyle [3]\;\;log2(3x + 2) - log4x = 3$
    Use a back-slash (\) before "log" . . . It looks better.
    Also you can make subscripts with _2 and _4

    We have: .$\displaystyle \log_2(3x+2) - \log_4(x) \:=\:3$

    Use the Base-change Formula on the second log:
    . . $\displaystyle \log_4(x) \:=\:\frac{\log_2(x)}{\log_2(4)} \:=\:\frac{\log_2(x)}{2}$

    The equation becomes: .$\displaystyle \log_2(3x+2) - \frac{1}{2}\log_2(x) \:=\:3$

    Multiply by 2: .$\displaystyle 2\log_2(3x+2) - \log_2(x) \:=\:6\quad\Rightarrow\quad \log_2(3x+2)^2 - \log_2(x) \:=\:6$

    Then we have: .$\displaystyle \log_2\left[\frac{(3x+2)^2}{x}\right] \:=\:6$

    Can you finish it now?



    $\displaystyle [4]\;\;(e^x + e^-x) / 2 = 3$

    Multiply by 2: .$\displaystyle e^x + e^{-x} \:=\:6$

    Multiply by $\displaystyle e^x\!:\;\;e^{2x} + 1 \:=\:6e^x\quad\Rightarrow\quad e^{2x} - 6e^x + 1\:=\:0$

    And we have a quadratic: .$\displaystyle \left(e^x\right)^2 - 6\left(e^x\right) + 1 \:=\:0$

    Proceed as we did in part [2].

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  5. #5
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    Quote Originally Posted by RedSpades View Post
    F. Solve each exponential equation.
    1. $\displaystyle 0.3(4^0.2x) = 0.2$
    2. $\displaystyle 9^x - 3^(x+1) + 1 = 0$
    3. $\displaystyle log2(3x + 2) - log4x = 3$
    4. $\displaystyle (e^x + e^-x) / 2 = 3$
    ...
    Hello,

    to 1.:

    $\displaystyle 0.3(4^{0.2x}) = 0.2~\iff~4^{0.2x} = \frac23~\iff~x=5\cdot \log_{4}\left(\frac23\right)$

    to 2.:

    $\displaystyle 9^x - 3^{x+1} + 1 = 0~\iff~3^{2x}-3\cdot 3^x+1=0$ Use $\displaystyle y=3^x$ to get $\displaystyle y^2-3y+1=0$. Solve for y. Don't forget to re-substitute to calculate x.

    to 3.:

    $\displaystyle \log_{2}(3x + 2) - \log_{4}(x) = 3~\iff~\log_{2}(3x+2)-\frac12 \cdot \log_{2}(x) = 3$$\displaystyle ~\iff~\log_{2}\left(\frac{3x+2}{\sqrt{x}}\right)=3$ Therefore $\displaystyle \frac{3x+2}{\sqrt{x}} = 8$ Solve for x.

    to 4.:

    $\displaystyle (e^x + e^{-x}) / 2 = 3~\iff~e^x+\frac1{e^x}=6~\iff~(e^x)^2+1=6e^x$ Use $\displaystyle y = e^x$ to get $\displaystyle y^2-6y+1=0$ . Solve for y. Don't forget to resubstitute to calculate x.
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