Find The Domain Of The Logarithmic Function

**J_Pettway** · March 31st 2008, 06:49 AM

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

**colby2152** · March 31st 2008, 06:51 AM

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.

$(x+9)^2 > 0$

$x+9>0$

$x>-9$

**Plato** · March 31st 2008, 09:28 AM

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 $\left\{ {x \in \mathbb{R}:x \ne - 9} \right\}$ .

**colby2152** · March 31st 2008, 09:52 AM

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 $\log _8 \left( {x + 9} \right)^2 = 2\log _8 (x + 9)$ ???

**Plato** · March 31st 2008, 10:02 AM

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.

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