# Thread: Domain of e function

1. ## Domain of e function

How do I find the domain of
$\displaystyle y=\sqrt{e^{5x}-2}$

I have a test in a few hours so thanks for any help I can get!

2. Can you have negatives inside a radical?

What values of x cause the argument of the radical to be zero or greater?

This will be your domain.

3. Hello, lil_cookie!

How do I find the domain of: .$\displaystyle y\:=\:\sqrt{e^{5x}-2}$

The radicand must be nonnegative.

. . $\displaystyle e^{5x} - 2 \:\geq \:0$

. . . . .$\displaystyle e^{5x} \:\geq \:2$

. . . . .$\displaystyle 5x \:\geq \:\ln 2$

. . . . . $\displaystyle x \:\geq \:\tfrac{1}{5}\ln2$

4. Thank you!

Here is another one:

$\displaystyle log_{4}+log_{4}(3x+11)=1$
$\displaystyle x(3x+11)=4$
$\displaystyle 3x^{2}+11x-4=0$
I get x=1 and x=-11

The q is to find all solutions
Is this correct, or did I go wrong?

5. Originally Posted by lil_cookie
Thank you!

Here is another one:

$\displaystyle log_{4}+log_{4}(3x+11)=1$
$\displaystyle x(3x+11)=4$
$\displaystyle 3x^{2}+11x-4=0$
I get x=1 and x=-11

The q is to find all solutions
Is this correct, or did I go wrong?
You didn't solve the quadratic correctly - use the quadratic formula to solve.

Also log terms must be greater than 0 so the domain is $\displaystyle x > -\frac{11}{3}$

6. Yes I was thinking I should have used the quad formula....thank you so much

7. Originally Posted by lil_cookie
Here is another one:

$\displaystyle log_{4}+log_{4}(3x+11)=1$
What is the argument of the first logarithmic term?