Solve the following logarithmic equations: (Criteria A)a) log4(2x+1)=2b) 3log(x−2)=6c) 4ln(2x+3)=4d) log3(x+1)−log3(2x−1)=1e) 2loga(3x+1)=loga(9)

Sure, let's solve these logarithmic equations one by one:

a) log4(2x+1)=2 To solve this, we can rewrite the equation in exponential form: 4^2 = 2x + 1 16 = 2x + 1 Subtract 1 from both sides: 15 = 2x Divide by 2: x = 15/2 = 7.5

b) 3log(x−2)=6 First, divide both sides by 3: log(x-2) = 2 Rewrite in exponential form: 10^2 = x - 2 100 = x - 2 Add 2 to both sides: x = 102

c) 4ln(2x+3)=4 First, divide both sides by 4: ln(2x+3) = 1 Rewrite in exponential form: e^1 = 2x + 3 e = 2x + 3 Subtract 3 from both sides: e - 3 = 2x Divide by 2: x = (e - 3)/2

d) log3(x+1)−log3(2x−1)=1 Combine the logs using the quotient rule: log3[(x+1)/(2x-1)] = 1 Rewrite in exponential form: 3^1 = (x+1)/(2x-1) 3 = (x+1)/(2x-1) Multiply both sides by (2x-1): 3(2x-1) = x + 1 6x - 3 = x + 1 Subtract x from both sides: 5x - 3 = 1 Add 3 to both sides: 5x = 4 Divide by 5: x = 4/5

e) 2loga(3x+1)=loga(9) First, divide both sides by 2: loga(3x+1) = loga(3^2) Since the bases and the logs are the same, the arguments must be equal: 3x + 1 = 3^2 3x + 1 = 9 Subtract 1 from both sides: 3x = 8 Divide by 3: x = 8/3

So, the solutions are x = 7.5, x = 102, x = (e - 3)/2, x = 4/5, and x = 8/3.