Solving a three-part log equation using the log laws The question asks:
Solve $$\log_5(x-1) + \log_5(x-2) - \log_5(x+6)= 0 $$
I know that according to log laws, addition with the same base is equal to multiplication and subtraction is equal to division (and vice versa)
By doing this I get $$\log_5{(x-1)(x-2)}=0
                                  (x+6)$$
By moving the $x+6$ to the other side it should become zero, ($x+6 \times 0$), however the answer in the textbook moves the $x+6$ to the other side of the equation and then solves. Why am I wrong? And when would you do my method then?
 A: By the same rules you mentioned you should get $$\log_5 \left[(x-1)(x-2)\right]=\log_5[x+6]$$ which after exponentiation with $5$ becomes $$(x-1)(x-2)=x+6$$
If you want to do it by including all three terms in the logarithm, you get the same:
$$\log_5 \left[\frac{(x-1)(x-2)}{x+6}\right]=0$$ but now you can't just multiply by $x+6$ on both sides and expect it to disappear in the denominator in the $\log$, since it is inside the $\log$. What you have to do is to exponentiate both sides to get $$\frac{(x-1)(x-2)}{x+6}=5^0=1$$
which can of course be rewritten as $$(x-1)(x-2)=x+6$$
So both methods lead to the same result. 
A: Laws of logarithm needed:
$$ \log_a pq = \log_a p + \log_a q$$
$$ \log_a \left(\frac{p}{q}\right) = \log_a p - \log_a q$$
$$\log_b a = 0  \ \implies a = b^0  $$
Using these three laws, your equation can be immediately reduced to:
$$ \log_5 \left(\frac{(x-1)(x-2)}{x+6}\right) = 0$$
Note: this is the step you got it wrong:
$$\left(\frac{(x-1)(x-2)}{x+6}\right) = 5^0 = 1 $$
$$ x^2 -3x +2 =x+6$$
A: or we have $$\log_5((x-1)(x-2))=\log_5(x+6)$$ thus we get $$(x-1)(x-2)=x+6$$
A: Well, the mistake in your approach is the use of the $\log$ properties.
Let's rewrite the initial equation as $\log_5\left((x-1)(x-2)\cdot \frac{1}{x+6}\right)=0.$
Then we have (since $\log_5 1=0$ and since $\log$ is $1-1$) that $(x-1)(x-2)\cdot \frac{1}{x+6}=1.$
Then, just do some calculations to arrive at the solution. Do not forget to take the restrictions $x-1>0,$ $x-2>0$ and $x+6>0$ (which are equivalent to $x>2$) in order the $\log_5$ to be well defined.
