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Common Logs

Key Questions

  • There are 2 ways.

    The math way is to understand how to convert bases:

    #log a=(ln a)/(ln 10)#

    The second way is to use the "CATALOG" button, "L", scroll down to "log" and press enter.

    Here is an example of #log 50#:

    #=(ln 50)/ln(10)#
    #~~1.69897#

  • Answer:

    See the explanation.

    Explanation:

    If you have technology available for the logarithm in some other base (#e# or #2#), use

    #log_10 n = log_b n / log_b 10# (where #b = e " or "2#)

    With paper and pencil, I don't know a good series for #log_10 n#.

    Probably the simplest way is to use a series for #ln n# and either a series or memorization for #ln 10 ~~ 2.302585093#

    For #ln n#, let #x=n-1# and use:

    #ln n = ln (1 + x) = x − x^2/2 + x^3/3- x^4/4+x^5/5- * * * #

    After you find #ln n#, use division to get #log_10 n ~~ ln n / 2.302585093#

  • Answer:

    The inverse of the function #f(x) = 10^x#

    Explanation:

    The function:

    #f(x) = 10^x#

    is a continuous, monotonically increasing function from #(-oo, oo)# onto #(0, oo)#

    graph{10^x [-2.664, 2.338, -2, 12.16]}

    Its inverse is the common logarithm:

    #f^(-1)(y) = log_10(y)#

    which as a result is a continuous, monotonically increasing function from #(0, oo)# onto #(-oo, oo)#.

    graph{log x [-1, 12.203, -1.3, 1.3]}

    Note that the exponential function satisfies:

    #10^a * 10^b = 10^(a+b)#

    Hence its inverse, the common logarithm satisfies:

    #log_10 xy = log_10 x + log_10 y#

Questions