Evaluate log_6(12)+(1/3)log_6(27)

Answering this question requires knowledge of the logarithm rules. Mainly how to manipulate logarithms.Firstly, notice that the multiplication in front of log_6(27) can be moved to be the power of the 27. Changing (1/3)log_6(27) to log_6(27^(1/3)). Taking the cubic root of 27 (x^3=27, what is x?) We see that this can then be simplified to (1/3)log_6(27)=log_6(3).The overall expression now becomes log_6(12)+log_6(3).We now use another logarithm rule, that when two logarithms with the same base and nothing multiplying either of them are added together, this is equivalent to log_a(x)+log_a(y)=log_a(xy). Therefore log_6(12)+log_6(3)=log_6(12*3)=log_6(36).Transforming this into exponential form, so we can see what this is equal to, we use the transformation formula:log_a(b)=c b=a^cSo, y=log_6(36) 36=6^yWhat power do we need to raise 6 to to get the answer as 36?2Therefore we have our solution.log_6(12)+(1/3)log_6(27)=2