Question

Can someone explain the change of base formula when evaluating logarithms? Thank you

Answer 1

The logarithm functinon is basically determining the value of x in:
y = b^x, where b is the base, or the subscript after the log, and y is the value in the parentheses.

With that, there isn't really an easy way to determine the value of x without using a little trial and error. The best way I found is to play around with the logarithm function and practice the ability to estimate them. For example, 15^1 = 15, so log15(15)=1, 5^6 = 625, so log5(625)=6, etc...

If you know the value of log_b(y), say x, then you can determine the value of y by calculating b^x.

Log7(y) = 3 => 7^3 = 343 => y=343.


And one more thing, an interesesting side note:
If your calculator can only calculate logarithms in base 10 and base e (ln), you can easily calculate logarithms in base b:

log_b(y) = log(y)/log(b), an interesting formula I used to play around with to help myself learn logarithms.

source: http://answers.yahoo.com/question/index?qid=20090308233845AA0ifnW

Answer 2

Suppose you have a log \[\log_9(27)\]The change of base formula allows you to turn this formula into a much nicer formula. Notice that \(9=3^2\) and \(27=3^3\). This means that changing into base 3 might be a good way to proceed. This means that we have \[\large \log_9(27)={\log_3(27) \over \log_3(9)}\]This is easy to evaluate as \[{\log_3(27) \over \log_3(9)}={3 \over 2}\]
In general, the change of base allows you to transform a log function as follows\[\large \log_b(a)={\log_c(a)\over \log_c(b)}\]Where \(a, b, c>0\). As we just saw, this can be a rather helpful tool.

Answer 3

Thank you so much guys! Your answers are so helpful. I really appreciate it!

Answer 4

You're welcome.