How do you solve #log_4 3 + 5 log_4 x = log_4 7#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Binayaka C. Mar 29, 2018 Solution: #x ~~1.185(3dp)# Explanation: #log_4 3 +5 log_4 x=log_4 7# or #log_4 3 + log_4 x^5=log_4 7# or #log_4 (3*x^5)=log_4 7# or # 3x^5=7 or x^5 =7/3 or x = root(5)(7/3)~~1.185(3dp)# Solution: #x ~~1.185(3dp)# [Ans] Answer link Related questions What is a logarithmic model? How do I use a logarithmic model to solve applications? What is the advantage of a logarithmic model? How does the Richter scale measure magnitude? What is the range of the Richter scale? How do you solve #9^(x-4)=81#? How do you solve #logx+log(x+15)=2#? How do you solve the equation #2 log4(x + 7)-log4(16) = 2#? How do you solve #2 log x^4 = 16#? How do you solve #2+log_3(2x+5)-log_3x=4#? See all questions in Logarithmic Models Impact of this question 1785 views around the world You can reuse this answer Creative Commons License