Question 396570
The domain is the set of possible input values. Usually "x" is the input variable so the domain is sometimes the set of possible values for x.<br>
When finding domains you want to make sure certain "no-no's" cannot happen. These "non-no's" include (but are not limited to):<ul><li>Zero denominators,</li><li>Negative radicands (expressions within a radical are called radicands) of even-numbered roots (square roots, 4th roots, 6th roots, etc.).</li><li>Zero or negative arguments to logarithms.</li></ul>
In expression (a), you have both an even-numbered root, the square root, and a logarithm. So you must ensure that the radicand of the square root, "x", can never be negative: {{{x >= 0}}}. And you must ensure that the argument of the logarithm, 10-x, can never be zero or negative (IOW it must be positive): {{{10-x > 0}}}. Putting these two together we have:
{{{x >= 0}}} and {{{10-x > 0}}}
Solving the second one for x we get:
{{{x >= 0}}} and {{{10 > x}}}
(Important tip: Always read inequalities like thees from where the variable is. That means we read the first one from left to right and the second one from right to left!) These inequalities tell use that x must be greater than or equal to 0 and less than (but not equal to) 10. This is the domain of your expression.<br>
For expression (b) I'm not sure if you really mean "eye-en" which means nothing to me or "el-en" which means natural logarithm. If you really meant "In" then I don't know how to find the domain since I don't know what "In" means or whether it has some limits on its arguments.<br>
If you mean "el-en", ln, then please be more careful in your posts. Tutors are less likely to help when problems are unclear.<br>
For ln(x)+ln(2-x) we have two logarithms. we must ensure that both arguemtns remain positive:
{{{x > 0}}} and {{{2-x > 0}}}
Solving the second one we get:
{{{x > 0}}} and {{{2 > x}}}
This tells us that x must be greater than (but not equal to) 0 and x must be less than (but not equal to) 2. This is the domain of expression (b) if they were ln's and not In's.