How do you express #log_50 23# in common logarithms?

1 Answer
Douglas K.
Oct 31, 2016

#log_50(23) = log_10(23)/log_10(50) = ln(23)/ln(50)#

Explanation:

Use the identity #log_c(a) = log_b(a)/log_b(c)#

The logarithms commonly found on calculators are base 10, #log_10 #, and base e, #ln#

#log_50(23) = log_10(23)/log_10(50) = ln(23)/ln(50)#

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