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Math Help - 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:

    lne^2

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

    y = e^x

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

    lne^3

    Answer:
    Quote Originally Posted by ThePerfectHacker View Post
    \ln e^n = n because e^{(n)} = e^n.
    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
    Last edited by RedSpades; October 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 y=e^x then  x = \ln y.

    And \ln e^n = n because 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.

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

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



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

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

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

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

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

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

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

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



    [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: . \log_2(3x+2) - \log_4(x) \:=\:3

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

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

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

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

    Can you finish it now?



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

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

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

    And we have a quadratic: . \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. 0.3(4^0.2x) = 0.2
    2. 9^x - 3^(x+1) + 1 = 0
    3. log2(3x + 2) - log4x = 3
    4. (e^x + e^-x) / 2 = 3
    ...
    Hello,

    to 1.:

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

    to 2.:

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

    to 3.:

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

    to 4.:

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