Domain of e function

**lil_cookie** · October 16th 2009, 08:23 AM

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

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

**stapel** · October 16th 2009, 08:30 AM

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.

**Soroban** · October 16th 2009, 08:31 AM

Hello, lil_cookie!

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

The radicand must be nonnegative.

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

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

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

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

**lil_cookie** · October 16th 2009, 08:48 AM

Thank you!

Here is another one:

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

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

**e^(i*pi)** · October 16th 2009, 10:12 AM

Originally Posted by lil_cookie

Thank you!

Here is another one:

$log_{4}+log_{4}(3x+11)=1$
$x(3x+11)=4$
$3x^{2}+11x-4=0$
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 $x > -\frac{11}{3}$

**lil_cookie** · October 16th 2009, 10:16 AM

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

**stapel** · October 16th 2009, 11:58 AM

Originally Posted by lil_cookie

Here is another one:

$log_{4}+log_{4}(3x+11)=1$

What is the argument of the first logarithmic term?

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