Question 639079
First of all, the argument of any logarithm must be positive. So
{{{x/2 > 0}}}
which resolves to
{{{x > 0}}}<br>
Second, the radicand of a square root must not be negative. So:
{{{ln(x/2) >= 0}}}
You might be able to logically dtermine from this that
{{{x/2 >= 1}}}
After all, an exponent of 0 results in 1. So any argument of 1 or greater will have an exponent greater than 0. If you can't see this then we have to solve {{{ln(x/2) >= 0}}}. We solve the inequality by rewriting it in exponential form:
{{{x/2 >= e^0}}}
Since any non-zero number, including e, to the zero power is 1 this becomes:
{{{x/2 >= 1}}}
Multiplying by 2 we get:
{{{x >= 2}}}<br>
From the fact that the argument of the log had to be positive we found that
{{{x > 0}}}
must be true. From the fact that radicands of square roots must bot be negative we found that
{{{x >= 2}}}
must be true. Since both of these must be true, the domain must be
{{{x >= 2}}}
(since any number greater than or equal to 2 must also be greater than 0).