How do you calculate #log_9(1/7)#? Precalculus Properties of Logarithmic Functions Common Logs 1 Answer A. S. Adikesavan May 30, 2016 #-loh 7/log 9=-0.8856#, nearly. Explanation: Use #log_b a^n=n log_b a and log_b a=log_c a/log_c b# Here, #log_9(1/7)=log_9(7^(-1))=-log_9 7= - log 7/log 9# 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 1261 views around the world You can reuse this answer Creative Commons License