How do you solve #log_5x + log_3 x=1#? Precalculus Properties of Logarithmic Functions Logarithm-- Inverse of an Exponential Function 1 Answer Shwetank Mauria May 7, 2018 #x=1.9211# Explanation: #log_5x+log_3 x=1# can be written as #logx/log5+logx/log3=1# or #log3logx+log5logx=log3log5# or #logx=(log3log5)/(log3+log5)# or #x=10^((log3log5)/(log15))# = #10^((0.4771*0.6990)/1.1761)# = #10^0.2836# = #1.9211# Answer link Related questions What is a logarithm? What are common mistakes students make with logarithms? How can a logarithmic equation be solved by graphing? How can I calculate a logarithm without a calculator? How can logarithms be used to solve exponential equations? How do logarithmic functions work? What is the logarithm of a negative number? What is the logarithm of zero? How do I find the logarithm #log_(1/4) 1/64#? How do I find the logarithm #log_(2/3)(8/27)#? See all questions in Logarithm-- Inverse of an Exponential Function Impact of this question 2786 views around the world You can reuse this answer Creative Commons License