Common Logs
Key Questions
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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#