Can you please answer those three question

1. 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.

2. 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.

3. 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.

Hi Pierre,

1. If x is greater than zero then the logarithm with base x is defined by
   log_x a = y if and only if x^y = a

   The reason that log_x is not defined for negative x is the expression x^y. If x is negative then x^1/2, x^1/4, x^3/4, x^1/8... are not real numbers.

2. 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.

3. Yes I think the dots you specify are multiplication. I think the expression is (writing * for multiplication)
   3*log_a r - (1/2)*log_a 5 + (1/3)*log_a t

   The crutial fact here is the rule of logarithms that says

   log_a x^y = y*log_a x

   Hence

   3*log_a r - (1/2)*log_a 5 + (1/3)*log_a t
   = log_a r³ - log_a 5^1/2 + log_a t^1/3

   Now use the other two rules of logarithms to express this as a single logarithm.

   Cheers,
   Harley