How do you use #log_10 5=.6990 and log_10 7=.8451# to evaluate the expression# log_10 (7/5)#? Precalculus Properties of Logarithmic Functions Common Logs 1 Answer A. S. Adikesavan May 28, 2016 0.1461 Explanation: Use the implied #log# for #log_10 and log (m/n)=log m-log n#. #log (7/5) = log 7 - log 5=0.8451-0.6990=0.1461#. Answer link Related questions What is the common logarithm of 10? How do I find the common logarithm of a number? What is a common logarithm or common log? What are common mistakes students make with common log? How do I find the common logarithm of 589,000? How do I find the number whose common logarithm is 2.6025? What is the common logarithm of 54.29? What is the value of the common logarithm log 10,000? What is #log_10 10#? How do I work in #log_10# in Excel? See all questions in Common Logs Impact of this question 2649 views around the world You can reuse this answer Creative Commons License