How to find the domain of a log function?

Question

So I have this function

$f(t)=\log(t^2+5t)$

And I want to find the domain.

Now I am taking into consideration $t^2+5$ which I know needs to be greater than $0$ and $t$ that needs to be greater than $0$. And until here im good.

On the answers sheet it says that $t>0$ and that's fine.

But then it says that $t+5>0$ when $t<-5$. And that for me doesn't make any sense whatsover. I take a number that is smaller than $t$ since that's what $t<-5$ implies. E.g $-7$. Then $-7+5$ is NOT a positive number and is not greater than $0$.

I thought that t+5>0 when t>-5. Then the domain would be from (-5,0) and from (0 to inifity).

I have no idea what I am not understanding , I have dyscalulia and any simple help would be great . Thanks!

The answer that the professor told me is (−∞, −5) ∪ (0, +∞).

Answer 1

If you want to find the domain of $f(t) = \log(t^2+5)$, the part $t^2+5$ needs to be bigger than zero, that's correct.

But for any number that you fill in for $t$, a positive number will be the outcome, because it is squared. ($5^2=25, (-5)^2=25$).

If you do $t^2+5$, still all values are positive, because any positive number$+5$ will be positive.

So the domain should be $<-\infty,\infty>$.

EDIT

The new equation is $f(t) = \log(t^2+5t)$.

Than for every $t$: $t^2+5t>0$. If you set $t^2+5t=0$ to find the intersection points, than $t(t+5)=0$ and you know that $t=0$ and $t=-5$.

If you fill in any point lower, in between and above those points, you can see how the equation works.

So pick for example $t=-6$, $t=-3$, $t=3$ and fill in.

$t=-6$ --> $(-6)^2+5\times -6 = 6$ which is bigger than zero so the formula works here.

$t=-3$ --> $(-3)^2+5\times -3 = -6$ which isn't bigger than zero so the formula doesn't work.

$t=3$ --> $(3)^2+5\times 3 = 24$ which is bigger than zero so the formula works here.

And than you know, that the domain is $(-\infty, -5)\cup(0, +\infty).$

Answer 2

We know that $\log t$ is defined for $t>0$. When you have $\log(\langle\textit{expression}\rangle)$, you have to set $$ \langle\textit{expression}\rangle>0 $$ in order to find where $\log(\langle\textit{expression}\rangle)$ is defined. In your case the condition becomes $$ t^2+5t>0 $$ Now we have to analyze this inequality; it factors as $t(t+5)>0$, which is satisfied when both factors are positive or both are negative.

The first case happens when $t>0$ (because in this case $t+5>0$ too).

The second case happens when $t+5<0$ (because in this case $t<0$ too.

Thus the domain is $(-\infty,-5)\cup(0,\infty)$.

It's simpler if you graph $y=x^2+5x$, which is a convex parabola meeting the $x$-axis at $-5$ and $0$:

This way you clearly see when $y>0$, that is, for $x<-5$ or $x>0$.

Note that it's not $t+5>0$ when $t<-5$, but $t(t+5)>0$.

Answer 3

$$f(t) = \log(t^2+5t)$$

You need the argument of the logarithm to be positive. Hence, you have

$$t^2+5t > 0$$

$$t(t+5) > 0$$

For a product to positive, either both factors are positive or both are negative. (This is an alternative to the other procedure posted.)

Case $1$:

$$t > 0 \quad t+5 > 0$$

$$t > 0 \quad t > -5$$

For $t > 0$, both conditions are true, so the factors are positive.

Case $2$:

$$t < 0 \quad t+5 < 0$$

$$t < 0 \quad t < -5$$

For $t < -5$, both conditions are true, so the factors are negative.

Hence, the domain becomes $(-\infty, -5)\cup(0, +\infty).$

Answer 4

We need to find domain of the function $log(t^2+5t)$ for this function to be defined $(t^2+5t)>0$ therefore we can write $t(t+5)$, this gives us two critical values ${-5,0}$, $-\infty<t<-5$ as well $t>0$ therefore answering in proper format $t$ lies in $(-\infty,-5)\cup(0,+ \infty)$