Find the value of logarithm of square root b to the base b - logarithm of 1/a to the base a?

Can anybody help me to solve this ?|||Logarithms and exponents have a fairly simple relationship:





1.) a^(log_base_a(b)) = b





or





2.) log_base_a(a^b) = b





Put in english, the log is the inverse operation of the exponent. If you take a log of a number in a base, you can "undo" the function by taking the exponent or vice-versa.





So for your specific problem:





log_base_b(sqrt(b)) = log_base_b(b^.5)


This matches equation (2) above, so we can say:


log_base_b(sqrt(b)) = log_base_b(b^.5) = .5





log_base_a(1/a) = log_base_a(a^-1) = -1 (also using equation 2)





Finally


log_base_b(sqrt(b)) - log_base_a(1/a) = .5 - (-1) = 1.5





Clear enough?|||The log of sqrt(b) to base b is 1/2 because b^(1/2) is the square root of b.



The log of 1/a to base a is -1 as a^-1 = 1/a



edit:



Oh sorry you are looking for the diference between the two:



(1/2)- (-1) = 1 + 1/2 or 3/2 or 1.5|||log[b] sqrt(b) = 1/2


log[a] 1/a = -1


So, the answer is 1/2 + 1 = 3/2