How do you find the value of #log_12 18# using the change of base formula?

1 Answer
Nov 1, 2016

Answer:

= #log_12 18 = 1.163#

Explanation:

Some calculators can work with different bases, but some only have #log_10#.
It is useful to understand where the change of base rule comes from.

Log form and index form are interchangeable.

#log_12 18 = x hArr 12^x = 18" "larr# x must be 1.??????

#color(white)(xxxxxxx)log_10 12^x = log_10 18" "larr# log both sides

#color(white)(xxxxxxx)xlog_10 12 = log_10 18" "larr# log power law

#color(white)(xxxxxxxxxxxx)x = (log_10 18)/(log_10 12)" "larr# solve for x

#color(white)(xxxxxxxxxxxx)x =1.2552725/1.0791812" "larr# calculate x

#color(white)(xxxxxxxxxxxx)x =1.1631712" "larr# calculate x

Now answer to whatever level of accuracy is required.
Usually 2 or 3 decimal places is sufficient.

This is called the "change of base rule" - the base has been changed from 12 in the given example to base 10 for the numerator and denominator. Now you can calculate the answer using any scientific calculator (or even find the values on a table)

#log_a b = (log_c b)/(log_c a)" "larr 'c'# is usually 10, can be any base