# Thread: Finding domain, solving exponential equation and finding factors.

1. ## Finding domain, solving exponential equation and finding factors.

Hi guys! I would really love help with some questions on my exam review. My exam is tomorrow and I'm completely confused! Help would be very much appreciated.

Find the domain of:
$f(x) = lnX / x^2 + 4$

Solve for X:
$2^x^-^1 = 3^x$

I know I need to make them the same base, but I don't know what would be simplest way to do that here.

Determine if (x - 1) is a factor of:
$-x^7^6^2 + 950x^3^4^8 - 938x^2^5 - 13$

I have absolutely no clue what to do here, except synthetic division with 762 places...

2. ## Re: Having some issues with exam review

For the first part, do you know how to find the domain of a function?

For the next, there are various approaches. What did you try? Here's one suggestion:

$2^{x-1}=3^x$

Taking logs:

$(x-1)\ln{2}=x\ln{3}$

$x\ln{2}-\ln{2}=x\ln{3}$

$x\ln{2}-x\ln{3}=\ln{2}$

$x(\ln{2}-\ln{3})=\ln{2}$

..etc.

For the last part, $x-1$ will be a factor if $f(1)=0$

3. ## Re: Having some issues with exam review

Usually for a rational I would just factor the denominator, but this isn't factorable by the conventional (x +/- a)(x +/- b)... method and I'm not sure if/how the log would affect the answer.

Ah I see, okay that makes sense.

How do you know, though? Is there some way of finding that out without using synthetic division?

4. ## Re: Having some issues with exam review

Originally Posted by Oddrick
Usually for a rational I would just factor the denominator, but this isn't factorable by the conventional (x +/- a)(x +/- b)... method and I'm not sure if/how the log would affect the answer.
If $x^2+4=0$, then $x^2=-4$. This cannot be the case for real values of $x$. The only thing restricting the domain is $\ln(x)$.

Originally Posted by Oddrick
Ah I see, okay that makes sense.

Originally Posted by Oddrick
How do you know, though? Is there some way of finding that out without using synthetic division?
Yes, using the factor theorem.

5. ## Re: Having some issues with exam review

So since
X = ln2/ln2-ln3

would that mean that
X = 1/-ln3

or
X = -ln3

This is probably a very dumb question. XP

6. ## Re: Having some issues with exam review

Originally Posted by Oddrick
So since
X = ln2/ln2-ln3

would that mean that
X = 1/-ln3

or
X = -ln3

This is probably a very dumb question. XP
Is that how you cancel terms in a fraction?!? Your instructor would be shocked.
$x=\frac{\ln{2}}{\ln{2}-\ln{3}}$

$x=\frac{\ln{2}}{\ln{\frac{2}{3}}}$

Now you could use the change of base rule to simplify.

7. ## Re: Having some issues with exam review

Wait, I'm confused as to how you got the
${\ln{\frac{2}{3}}}$
I missed pretty much all the discussion about logs so this is very foreign to me...

And for the one about factor theorem, I got a remainder of -2 so it would be no, right? Does that seem right?

8. ## Re: Having some issues with exam review

Originally Posted by Oddrick
Wait, I'm confused as to how you got the
${\ln{\frac{2}{3}}}$
I missed pretty much all the discussion about logs so this is very foreign to me...

And for the one about factor theorem, I got a remainder of -2 so it would be no, right? Does that seem right?
You're right about the factor/remainder theorem.

However, it's very obvious that you have very little understanding when it comes to the concept of logarithms. You need to catch up on missed work quickly rather than on the night before your exam. You require more help than you can receive in a quick forum session. I suggest finding an online review of this topic, which you'll need to study for a few hours until you have a thorough understanding.

9. ## Re: Having some issues with exam review

can you explain it

Find the domain of:
f(x) = lnX / x^2 + 4

10. ## Re: Having some issues with exam review

It's supposed to be $f(x)=\frac{\ln{x}}{x^2+4}$, and as $x^2+4\neq{0}$ for real $x$, the only thing restricting the domain is $\ln(x)$, which can only take values for $x>{0}$

thank you =)