How do you solve #log_b(5)=0.558#?

1 Answer
May 3, 2016

Answer:

I found #b=e^2.8843=17.89~~18#

Explanation:

I would try changing base to your log choosing the natural log, i.e., #ln#:

#(ln(5))/(ln(b))=0.558#
so rearranging:
#ln(b)=(ln(5))/(0.558)=2.8843#
now we apply the definition of log to get:
#b=e^2.8843=17.89~~18#