How do you solve log(y) = 1/4 log16 + 1/2 log49log(y)=14log16+12log49? 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 14logy=14log16+12log49=log2+log7=log14 So y = 14y=14 Explanation: If a > 0a>0 then log a^b = b log alogab=bloga If a, b > 0a,b>0 then log ab = log a + log blogab=loga+logb So: log y = 1/4 log 16 + 1/2 log 49 = 1/4 log 2^4 + 1/2 log 7^2logy=14log16+12log49=14log24+12log72 = 1/4*4 log 2 + 1/2 * 2 log 7 = log 2 + log 7 = log (2*7)=14⋅4log2+12⋅2log7=log2+log7=log(2⋅7) = log 14=log14 That is: log y = log 14logy=log14 If log a = log bloga=logb then a = ba=b So y = 14y=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)=819x−4=81? How do you solve logx+log(x+15)=2logx+log(x+15)=2? How do you solve the equation 2 log4(x + 7)-log4(16) = 22log4(x+7)−log4(16)=2? How do you solve 2 log x^4 = 162logx4=16? How do you solve 2+log_3(2x+5)-log_3x=42+log3(2x+5)−log3x=4? See all questions in Logarithmic Models Impact of this question 3036 views around the world You can reuse this answer Creative Commons License