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Help with basic log stuff

  1. Sep 20, 2006 #1
    Here are some log questions that I don't really have any idea how to do. I'd appreciate any hints/help... (solve for x in all questions)

    1. [itex]2^x + 4^x = 8^x [/itex]
    All I could think of doing was converting them to the same base. I don't see how to simplify the left side of the equation so I can equate the exponents (if that's what I have to do). If I take the log of both sides I just end up getting [itex] x log {2} + x log {4} = log {8} [/itex], and I don't know where to go from there.

    2. [itex] log_x {2} = log_2x {8} [/itex]
    I really have no idea how to approach this.

    3. solve for x in terms of y
    [itex] y = (e^2x - e^-2x) / 2 [/itex]
    Again, no idea. (specifically, I don't know how to take the natural logarithm of both sides of this equation which is what i think i need to do)

    I need some help with this log/ln stuff... if nothing else, are there any good links I can read up on? I forgot pretty much everything I've learned about this stuff.

    Thanks
    Preet
     
  2. jcsd
  3. Sep 20, 2006 #2
    For 1, note that you can rewrite it as

    [tex]2^x + (2^x)^2 = (2^x)^3.[/tex]

    See if that helps!

    I'm going to assume that 2 is actually to solve

    [tex]\log_x2 = \log_{2x}8.[/tex]

    Try using [tex]\log_a b = \frac{\log_c b}{\log_c a}[/tex] on the RHS to convert all the logs to the same base.

    I'm also going to assume 3 is actually to solve

    [tex]y = \frac{e^{2x}-e^{-2x}}{2}[/tex]

    for x in terms of y.

    Can you solve

    [tex]y = \frac{u + \frac{1}{u}}{2}[/tex]

    for u in terms of y? If so, you should be able to do your question too!
     
    Last edited: Sep 20, 2006
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