# Thread: Find The Domain Of The Logarithmic Function

1. ## Find The Domain Of The Logarithmic Function

f(x)= log8 (x + 9)^2

2. Originally Posted by J_Pettway
f(x)= log8 (x + 9)^2
You can only take the log of positive numbers, so $\displaystyle (x+9)^2$ must be greater than zero.

$\displaystyle (x+9)^2 > 0$

$\displaystyle x+9>0$

$\displaystyle x>-9$

3. Originally Posted by colby2152
You can only take the log of positive numbers, so $\displaystyle (x+9)^2$ must be greater than zero.
$\displaystyle (x+9)^2 > 0$, $\displaystyle x+9>0$, $\displaystyle x>-9$
Because $\displaystyle \log _8 \left( {x + 9} \right)^2 = 2\log _8 \left| {x + 9} \right|$
The domain is $\displaystyle \left\{ {x \in \mathbb{R}:x \ne - 9} \right\}$.

4. Originally Posted by Plato
Because $\displaystyle \log _8 \left( {x + 9} \right)^2 = 2\log _8 \left| {x + 9} \right|$
The domain is $\displaystyle \left\{ {x \in \mathbb{R}:x \ne - 9} \right\}$.
Wouldn't log properties say $\displaystyle \log _8 \left( {x + 9} \right)^2 = 2\log _8 (x + 9)$ ???

5. Originally Posted by colby2152
Wouldn't log properties say $\displaystyle \log _8 \left( {x + 9} \right)^2 = 2\log _8 (x + 9)$ ???
No.
Realize that $\displaystyle \left( {x + 9} \right)^2 = \left| {x + 9} \right|^2$.

Of course, your answer is correct if the problem means $\displaystyle \left[ {\log _8 \left( {x + 9} \right)} \right]^2.$
That is the way most computer algebra systems would read it.

But I don't think that is what was meant.