How do I find the derivative of #(x)=Log_5(x)#?

1 Answer
May 22, 2018

#d/dx log_5(x) = 1/(xln(5)) = log_5(e)/x#

Explanation:

Recall from logarithmic properties that #log_5(x) = ln(x)/ln(5)#

Thus #d/dx log_5(x) = d/dx ln(x)/ln(5)#

Since #1/ln(5)# is a constant,

#d/dx ln(x)/ln(5) = 1/ln(5) d/dx ln(x) = 1/ln(5) * 1/x = 1/(xln5)#

You can also use the fact that #1/ln5 = log_5(e)# to present the answer differently

#d/dx log_5(x) = 1/(xln5) = log_5(e)/x#