Given #log_b 2= 0.3562#, #log_b 3=0.5646#, and #log_b 5=0.8271#, how do you evaluate #log_b15#?

1 Answer
May 13, 2016

Answer:

#log_b15=1.3917#

Explanation:

To calculate #log_b15# we use the formula #log_ba*c=log_ba+log_bc#

First we have to write #15# as a product of numbers #2,3 and 5#. Number #2# will not be used because #15# is an odd number, but we can write that #15=3*5#

Now we can use the formula above to write the logarythm as a sum of 2 given values.

#log_b15=log_b3*5=log_b3+log_b5=0.5646+0.8271=1.3917#