How do you approximate #log_5 0.5# given #log_5 2=0.4307# and #log_5 3=0.6826#?
1 Answer
Oct 19, 2016
Explanation:
We will use the following:
#log(1) = 0# #log(a^x) = xlog(a)# #1/a = a^(-1)#
With that:
#=log_5(2^(-1))#
#=-1*log_5(2)#
#=-log_5(2)#
#~~-0.4307#
