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Math Help - Logarithmic Functions

  1. #1
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    Logarithmic Functions

    I dont understand how log(2)(2) = 1. Inserting it in my calculator returns ~.6 but my textbooks says it equals 1. Is it just some sort of fact I have to believe, or is there a way to determine it?
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  2. #2
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    Quote Originally Posted by ceasar_19134 View Post
    I dont understand how log(2)(2) = 1. Inserting it in my calculator returns ~.6 but my textbooks says it equals 1. Is it just some sort of fact I have to believe, or is there a way to determine it?
    The definition of \log to the base b of a is the power that b must be raised to
    to give a. So as b^1=b, we have \log_b(b)=1 for all positive bases b.

    RonL
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    Im still having trouble understanding. Perhaps it would simply be easier to explain how to go about solving log(2)(2) or log(2)(3)?

    The book doesnt explain at all- It simply shows of table of values when x = 2, 4, 8, 16, ..., 128.
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    Your problem may well be in notation. The way you write the problem.
    Here is an example.
    \begin{array}{l} Log_b \left( a \right) = x\quad  \Leftrightarrow \quad b^x  = a \\  Log_3 \left( {81} \right) = 4\quad  \Leftrightarrow \quad 3^4  = 81 \\  \end{array}
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    Didnt get very far before I got stuck again-

    "The base-10 logarithm is called the common logarithm. The common logarithm, log(10)x, is usually written as log x."

    I dont understand what happens to the 10. The equation could be arranged into 10^y=x, but that doesnt help me any, if anything its causes more confusion.
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ceasar_19134 View Post
    I dont understand how log(2)(2) = 1. Inserting it in my calculator returns ~.6 but my textbooks says it equals 1. Is it just some sort of fact I have to believe, or is there a way to determine it?
    Quote Originally Posted by ceasar_19134 View Post
    Didnt get very far before I got stuck again-

    "The base-10 logarithm is called the common logarithm. The common logarithm, log(10)x, is usually written as log x."

    I dont understand what happens to the 10. The equation could be arranged into 10^y=x, but that doesnt help me any, if anything its causes more confusion.
    I have an idea of what happened initially anyway.

    You plugged log(2)(2) into your calculator and got 0.60206, didn't you?

    The question your book is asking about is log_2(2), which is a matter of notation, as Plato suggested.

    Your calculator doesn't have a log button that will do log_2.

    What you did was
    log (2)(2) = log_{10}(2 \cdot 2) = log_{10}(4) \approx 0.60206

    So let's rethink your original problem. Set
    x = log_2(2)

    According to Plato's post this is equivalent to:
    2^x = 2

    What x value(s) satisfy this equation? The one that should immediately come to mind is x = 1. And that's the only solution, in fact.

    Thus
    x = log_2(2) = 1

    There is a way to get the calculator to work out your problem. I'm not going to tell you explicitly for fear of confusing you more than you already are. The name of the equation is the "Change of base formula." Look it up when you get a little more confidence using logarithms.

    -Dan
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    Perhaps I explained it wrong or am still managing not to see what is under my nose, but I now understand what I had done wrong with log"(2)"2.

    However, when presented with log"(10)"x = log x I cant advance- Rearranged yields 10^y=x, which doesnt make any sense to me.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ceasar_19134 View Post
    Perhaps I explained it wrong or am still managing not to see what is under my nose, but I now understand what I had done wrong with log"(2)"2.

    However, when presented with log"(10)"x = log x I cant advance- Rearranged yields 10^y=x, which doesnt make any sense to me.
    Ah, my apologies. I had misunderstood the post.

    We commonly write just "log" when we mean log_{10}.

    So your expression is using two different conventions, which is why you are confused. log(x) and log_{10}(x) are meant to be the same thing.

    (Warning: There are a number of members on the site, especially the European ones I think, that will tell you that log(x) is the same as log_e(x) where e = 2.71828.... All I can say is that conventions vary and that your calculator almost certainly uses the convention that log(x) = log_{10}(x).)

    -Dan
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