1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Evaluating Logarithms

  1. Oct 24, 2009 #1
    Hello

    I was wondering if someone could help me.

    I've got a question on evaluating logarithms, but I've not done anything like this before. I'm only used to the basic logarithmic stuff, and I still find that a bit confusing.

    So, I've been trying to do this question for hours now, and keep getting x=1/2, x = -2 or 1 = 1, despite knowing that x = 1

    Could anyone shed some light on this for me? I'm so confused.

    Thank you

    1. The problem statement, all variables and given/known data

    [tex]Evaluate : 2^x - 2^{1-x} = 1[/tex]


    2. Relevant equations

    [tex]logb^n = nlogb[/tex]


    3. The attempt at a solution

    [tex]2^x - 2^{1-x} = 1[/tex]

    [tex]log(2^x) - log(2^{1-x}) = log(1)[/tex]

    [tex]xlog(2) - (1-x)log(2) = 0[/tex]

    [tex]log(2)(x - (1-x)) = 0[/tex]

    [tex]log(2)(2x-1) = 0[/tex]

    [tex]2xlog(2) - 1log(2) = 0[/tex]

    [tex]2xlog(2) = log(2)[/tex]

    [tex]2x = \frac{log(2)}{log(2)}[/tex]

    [tex]2x = 1[/tex]

    [tex]x = \frac{1}{2}[/tex]
     
  2. jcsd
  3. Oct 24, 2009 #2
    If you take log on both sides, you get [tex]log(2^x - 2^{1-x}) [/tex] on the LHS, which is not equal to [tex]log(2^x) - log(2^{1-x})[/tex].
    The method to solve it is not by logarithms; try simplifying the expression to see if you can see the underlying equation present.
     
  4. Oct 24, 2009 #3

    Mark44

    Staff: Mentor

    Your next step is wrong. You can take the log of each side of an equation, but there is no property that says log(a - b) = log a - log b. That is what you've done, and it is invalid.

    Rewrite your equation as 2x - 2/2x = 1, and then multiply both sides by 2x. That should give you an equation that is quadratic in form (in powers of 2x) that you can solve.
     
  5. Oct 24, 2009 #4
    Thanks for the replies. :smile:

    I'd been doing questions with logarithms just before I got to this question, so when I saw the power of x I thought of logarithms and nothing else.

    Sooooo, how about this?

    [tex]2^x - 2^{1-x} = 1[/tex]

    [tex]2^x - 2^1 \cdot 2^{-x} = 1[/tex]

    [tex]2^x - \frac{2^1}{2^x} = 1[/tex]

    [tex]2^x - \frac{2}{2^x} - 1= 0[/tex]

    [tex]y = 2^x[/tex]

    [tex]y - \frac{2}{y} - 1 = 0[/tex]

    [tex]y^2 - 2 - y = 0[/tex]

    [tex]y^2 - y - 2 = 0[/tex]

    [tex]y^2 - 2y + 1y - 2 = 0[/tex]

    [tex]y(y - 2) + 1(y - 2) = 0[/tex]

    [tex](y - 2)(y + 1) = 0[/tex]

    [tex]y = 2[/tex] and [tex]-1[/tex]

    [tex]y = -1[/tex]

    [tex]2^x \neq -1[/tex] ([tex]2^x[/tex] cannot be negative)

    [tex]y = 2[/tex]

    [tex]2^x = 2[/tex]

    [tex]log(2^x) = log(2)[/tex]

    [tex]xlog(2) = log(2)[/tex]

    [tex]x = \frac{log(2)}{log(2)}[/tex]

    [tex]x = 1[/tex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Evaluating Logarithms
  1. Logarithm's and Such (Replies: 12)

  2. Evaluating logarithms (Replies: 4)

Loading...