# 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 $(x+9)^2$ must be greater than zero.

$(x+9)^2 > 0$

$x+9>0$

$x>-9$

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

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

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

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

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