Question 383943
{{{log((sqrt(100y)))}}}
To start we need to remember that a square root is the same as an exponent of 1/2. So we can rewrite your expression as:
{{{log(((100y)^(1/2)))}}}
Next we can use a property of logarithms, {{{log(a, (p^q)) = q*log(a, (p))}}}, to move the exponent of the argument out in front:
{{{(1/2)log((100y))}}}
Next we can use another property of logarithms, {{{log(a, (p*q)) = log(a, (p)) + log(a, (q))}}}, to rewrite the log of the product as the sum of the logs of the factors:
{{{(1/2)(log((100))+log((y)))}}}
Note the use of parentheses. It is an extremely good habit to use parentheses when substituting. In this case it helps us know that the Distributive Property applies with the 1/2:
{{{(1/2)log((100))+(1/2)log((y))}}}
Since the base of log is 10 and since {{{100 = 10^2}}}, {{{log((100)) = 2}}}:
{{{(1/2)(2)+(1/2)log((y))}}}
which simplifies to:
{{{1+(1/2)log((y))}}}