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Off-topic / Who wants to help me D:

  1. Nachoaddict
    Date: Sun, Feb 21 2010 01:59:50

    Help me solve this through natural logarithms for all of you who have taken Algebra 2 or are taking it?

    2e^(x-2) = e^(x) + 7

    I will wub you forever if you help me

  2. Kaffatsum
    Date: Sun, Feb 21 2010 02:08:04

    I just got with this chapter in Alg II (H).
    Let me try ..

  3. Clyde
    Date: Sun, Feb 21 2010 02:18:56

  4. Nachoaddict
    Date: Sun, Feb 21 2010 02:23:41

    QUOTE (Clyde @ Feb 20 2010, 06:18 PM) <{POST_SNAPBACK}>

    Spoiler:


    Where in the hell do you get pi and i?

  5. Kaffatsum
    Date: Sun, Feb 21 2010 02:46:39

    damn that was a confusing problem.
    x = ~5.946 <--- rounded
    x = 4 + ln 7 <---- exact

    2e^x-2 = e^x +7 <---------- original problem
    (x-2)(2) ln e = x ln e + ln 7 <- natural log both sides
    (x-2)(2) = x + ln 7 <---------- natural log of e cancels
    2x-4 = x + ln 7 <-------------- distributive property on left side of problem.. (2)(x-2)
    -4 - ln 7 = x -2x <------------- subtract ln 7 and subtract 2x
    -4 - ln 7 = -x <---------------- simplify right side of problem.. x-2x = -x
    4 + ln 7 = x <------------------ multiply everything by -1
    x = 4 + ln 7 <------------------ switch sides.

    EXTRA

    x = 4 + 1.946 <---------------- ln 7 = ~1.946 (rounded to the thousandths)
    x = ~5.946 <------------------- add and rounded.

  6. Nachoaddict
    Date: Sun, Feb 21 2010 03:04:22

    QUOTE (Kaffatsum @ Feb 20 2010, 06:46 PM) <{POST_SNAPBACK}>
    damn that was a confusing problem.
    x = ~5.946 <--- rounded
    x = 4 + ln 7 <---- exact

    2e^x-2 = e^x +7 <---------- original problem
    (x-2)(2) ln e = x ln e + ln 7 <- natural log both sides
    (x-2)(2) = x + ln 7 <---------- natural log of e cancels
    2x-4 = x + ln 7 <-------------- distributive property on left side of problem.. (2)(x-2)
    -4 - ln 7 = x -2x <------------- subtract ln 7 and subtract 2x
    -4 - ln 7 = -x <---------------- simplify right side of problem.. x-2x = -x
    4 + ln 7 = x <------------------ multiply everything by -1
    x = 4 + ln 7 <------------------ switch sides.

    EXTRA

    x = 4 + 1.946 <---------------- ln 7 = ~1.946 (rounded to the thousandths)
    x = ~5.946 <------------------- add and rounded.


    Lol that's what i got but it's not right. plug it into ur calculator?

  7. Kaffatsum
    Date: Sun, Feb 21 2010 03:13:03

    Pretty sure thats right. i did it like 5 times tongue.gif
    How do you know its not right?

  8. :DDD
    Date: Sun, Feb 21 2010 03:18:38

    read http://answers.yahoo.com/question/index?qi...11173757AARNpWS cuz they asked the same question lawl

  9. Kaffatsum
    Date: Sun, Feb 21 2010 03:22:30

    QUOTE (:DDD @ Feb 20 2010, 10:18 PM) <{POST_SNAPBACK}>
    read http://answers.yahoo.com/question/index?qi...11173757AARNpWS cuz they asked the same question lawl


    interesting..
    i think i see what i did wrong now.. sad.gif
    well i did get like a C on that chapter tongue.gif

  10. :DDD
    Date: Sun, Feb 21 2010 03:24:36

    lol im only in algebra 1b but i was bored so i googled the problem and there it was. lol

  11. Rorix
    Date: Sun, Feb 21 2010 07:28:14

    If you still need help:

    2e^(x-2) = e^x + 7
    2e^(x-2) - e^x = 7
    2(e^x)/(e^2) - e^x = 7
    e^x ( 2/e^2 - 1 ) = 7
    e^x = 7 / ( 2/e^2 - 1 )
    x = ln (7 / ( 2/e^2 - 1 ) )

    But 7 / ( 2/e^2 - 1 ) < 0 so no real solutions.

  12. Nachoaddict
    Date: Sun, Feb 21 2010 18:19:13

    QUOTE (Rorix @ Feb 20 2010, 11:28 PM) <{POST_SNAPBACK}>
    If you still need help:

    2e^(x-2) = e^x + 7
    2e^(x-2) - e^x = 7
    2(e^x)/(e^2) - e^x = 7
    e^x ( 2/e^2 - 1 ) = 7
    e^x = 7 / ( 2/e^2 - 1 )
    x = ln (7 / ( 2/e^2 - 1 ) )

    But 7 / ( 2/e^2 - 1 ) < 0 so no real solutions.


    Yeah I found out yesterday that this should be the right answer. I didn't know that 2e^(x-2) = (2e^x)/e^2

    It was in e format so I didn't compare it to the properties of log lol...

    Thanks everyone for your efforts smile.gif