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Homework Help: Prove that the logarithmic function is continuous on R.

  1. Jul 29, 2017 #1
    1. The problem statement, all variables and given/known data

    Prove that [itex]f\left(x\right)=\log_{a}x[/itex] is continuous for all [itex]\mathbb{R}[/itex].

    2. Relevant equations

    I must find a [itex]\delta>0\in\mathbb{R}[/itex] for a given [itex]\varepsilon>0[/itex]
    such that

    3. The attempt at a solution.

    I tried to use a direct proof solving [itex]\left|\log_{a}x-\log_{a}x_{0}\right|<\varepsilon[/itex] for [itex]x[/itex]. But this gives rise to a pair of values: [itex]\delta=x_{0}\left(a^{\varepsilon}-1\right)[/itex] and [itex]\delta=x_{0}\left(a^{-\varepsilon}-1\right)[/itex]. When I use them to build [itex]\left|x-x_{0}\right| < \delta[/itex] from the inequality
    I see myself in big trouble, as there is no way to generate [itex]\left|\log_{a}x-\log_{a}x_{0}\right| < \varepsilon[/itex].

    Can someone give me a hint, or a new strategy? I have searched for help in other forums, but hints are very sophisticated to follow.
    Thanks in advance.
  2. jcsd
  3. Jul 29, 2017 #2


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    How do you expect a function that is not defined at ##x =0## to be continuous on the entire real line?
  4. Jul 29, 2017 #3
    Oh, sorry, I forgot it! Continuous for all [itex] x>0 [/itex].
  5. Jul 29, 2017 #4


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    Why don't you try and see where ##\log x - \log x_0 = \log \frac{x}{x_0}## gets you? Those proofs can often be done by writing it the wrong way and start with ##|f(x)-f(x_0)| < \varepsilon## to obtain a condition for ##\delta## to get an idea and then turn the estimations, resp. conclusions around.
  6. Aug 1, 2017 #5


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    You have solved [tex]
    -\epsilon < \log_a (x/x_0) < \epsilon[/tex] correctly to get [tex]
    x_0(a^{-\epsilon} - 1) < x - x_0 < x_0(a^{\epsilon} - 1),[/tex] but the interval is not symmetric about zero.

    In this situation the closest end point to zero gives you your [itex]\delta[/itex]. For suppose [itex]-b < x - x_0 < c[/itex] for strictly positive [itex]b[/itex] and [itex]c[/itex]. If [itex]b < c[/itex] then [itex]|x - x_0| < b[/itex] implies [itex]-b < x - x_0 < b < c[/itex] as required; alternatively if [itex]b > c[/itex] then [itex]|x - x_0| < c[/itex] implies [itex]-b < -c < x - x_0 < c[/itex], again as required.
  7. Aug 1, 2017 #6
    Thank you very much, pasmith! I was so close to the answer!! Now I understand. Thank you very much.
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