Question 202923
There are several properties of logarithms which are useful when you want to manipulate expressions involving them:<ol><li>{{{log(a, (x*y)) = log(a, (x)) + log(a, (y))}}}
Used from left to right, this property can be used to separate factors in the argument of a logarithm into separate logarithms. Used from right to left this can be used to combine the sum of two logarithms into a single, equivalent logarithm.</li><li>{{{log(a, (x/y)) = log(a, (x)) - log(a, (y))}}}
Used from left to right, this property can be used to separate the numerator and denominator of a fraction in the argument of a logarithm into separate logarithms. Used from right to left this can be used to combine the difference of two logarithms into a single, equivalent logarithm.</li><li>{{{log(a, (x^y)) = y*log(a, (x))}}}
Used from left to right, this property can be used to "move" of the argument of a logarithm out in front of the logarithm (as a coefficient. Used from right to left this can be used to "move" a coefficient of a logarithm into the arguments as the exponent of the logarithm.</li><li>{{{log(a, (x)) = (log(b, (x)))/(log(b, (a)))}}}
This property is used most used from left to right in order to change the base of a logarithm from "a" to "b".</li></ol>
We will use these properties to simplify
{{{log(a,(a/sqrt(x))) - log(a, sqrt(ax))}}}
(If the above is not a correct representation of your problem, please repost the question using more parentheses to clarify "what goes with what and where".)
The argument of the first log is a fraction and we will use property #2 above to separate the numerator and denominator into separate logs:
{{{log(a, (a)) - log(a, (sqrt(x))) - log(a, (sqrt(ax)))}}}
{{{log(a, (a)) = 1}}}. 
And square roots are the same as an exponent of (1/2). Using these facts to rewrite our expression we get:
{{{1 - log(a, (x^(1/2))) - log(a, ((ax)^(1/2)))}}}
Now we can use property #3 above to "move" the exponents out from the arguments to the front of the 2ns and 3rd logs:
{{{1 - (1/2)log(a, (x)) - (1/2)log(a, (ax))}}}
Now we can use property #1 above to separate the "a" and the "x" in the third log into separate logs. (Note the parentheses used when substituting. They are critical.)
{{{1 - (1/2)log(a, (x)) - (1/2)(log(a, (a)) + log(a, (x)))}}}
Again {{{log(a, (a)) = 1}}}. 
{{{1 - (1/2)log(a, (x)) - (1/2)(1  + log(a, (x)))}}}
Simplifying we get:
{{{1 - (1/2)log(a, (x)) - (1/2) - (1/2)log(a, (x))}}}
Combining like terms (1 - (1/2) = 1/2; -1/2log - 1/2log = -1log):
{{{1/2 - log(a, (x))}}}