How do you solve #log(y) = 1/4 log16 + 1/2 log49#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer George C. Aug 2, 2015 #log y = 1/4 log 16 + 1/2 log 49 = log 2 + log 7 = log 14# So #y = 14# Explanation: If #a > 0# then #log a^b = b log a# If #a, b > 0# then #log ab = log a + log b# So: #log y = 1/4 log 16 + 1/2 log 49 = 1/4 log 2^4 + 1/2 log 7^2# #= 1/4*4 log 2 + 1/2 * 2 log 7 = log 2 + log 7 = log (2*7)# #= log 14# That is: #log y = log 14# If #log a = log b# then #a = b# So #y = 14# 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 2794 views around the world You can reuse this answer Creative Commons License