Subject: logarithms Can you please answer those three question inverse or the exponential function. logx 81 = 4, x expo4 = 81, x = +or- 4square root 81, x = +or- 3 s.s = {3} Why is -3 not acceptable. log3 ( x square - 1) = 1, 3expo 1 = x square - 1 x square = 4 x = +or- square root 4 x = +or- 2 s.s = {2,-2} Why is - 2 acceptable here. Express as a single logarithm. 3.loga r - 0.5.loga 5 + 0.33.loga t Does the dot after 3&.5&.33 means to multiply, if so how do I proceed. If not what does it means. Thank you Pierre. Hi Pierre, If x is greater than zero then the logarithm with base x is defined by logx a = y if and only if xy = a The reason that logx is not defined for negative x is the expression xy. If x is negative then x1/2, x1/4, x3/4, x1/8... are not real numbers. When you solve an equation, as you have done in this problem, you should, if possible, verify that the "solutions" you get are actually solutions to the original problem. For this problem log3 ( x2 - 1) = 1 substituting x = 2 or x = -2 on the left side gives log33 which is 1. Thus x = 2 and x = -2 are both solutions. Yes I think the dots you specify are multiplication. I think the expression is (writing * for multiplication) 3*loga r - (1/2)*loga 5 + (1/3)*loga t The crutial fact here is the rule of logarithms that says loga xy = y*loga x Hence 3*loga r - (1/2)*loga 5 + (1/3)*loga t = loga r3 - loga 51/2 + loga t1/3 Now use the other two rules of logarithms to express this as a single logarithm. Cheers, Harley Go to Math Central