Find all the real-number roots of the equation. Give an exact expression for the root and also a calculator approximation rounded to three decimal places.
log3 x + log3(x + 8) = 0

Question

Find all the real-number roots of the equation. Give an exact expression for the root and also a calculator approximation rounded to three decimal places. 
log3 x + log3(x + 8) = 0

anonymous · Answer

$$\log_3(x)+\log_3(x+8)=0$$

$$\log_3(x(x+8))=0$$
$$3^0=x(x+8)$$ 
$$x(x+8)=1$$

myininaya · Answer

$$\log_3(x(x+8))=0$$
3^                  3^
$$x(x+8)=1$$
$$x^2+8x-1=0$$
$$x=\frac{-8 \pm \sqrt{64-4(1)(-1)}}{2}=\frac{-8 \pm \sqrt{68}}{2}$$

anonymous · Answer

$$x^2+8x-1=0$$
$$x=-4+\sqrt{17}$$

anonymous · Answer

what is this strange formula i see above?

myininaya · Answer

but there is only one x

anonymous · Answer

yes indeed there is only one x, right between w and y

myininaya · Answer

since one is will make the inside of the log above negative

anonymous · Answer

$$\color{pink}{\text{you are getting sleepy... very sleepy}}$$

myininaya · Answer

$$x=\frac{-8 + \sqrt{68}}{2}$$
i trust you can simplify

anonymous · Answer

@tky myininaya got the same answer.  there are two roots of 
$$x^2+8x-1=0$$ but one is negative and since the domain log is positive numbers you cannot use the negative solution

anonymous · Answer

and yes, i can simplify 
$$\frac{-8+\sqrt{68}}{2}$$ by never getting it to begin with 
$$x^2+8x-1=0$$
$$x^2=8x-1$$
$$(x+4)^2-1+16=17$$
$$x+4=\pm\sqrt{17}$$
$$x=-4\pm\sqrt{17}$$

anonymous · Answer

typos galore, which is what i get for being a wise guy.  should be 
$$x^2+8x=1$$
$$(x+4)^2=1+16=17$$
etc