Solve the logarithmic equation: $\log_2(x^2 − 5x − 28) = 3$ I distributed the the log 
$$\log_2 x^2 - \log_2 2x - \log_2 27 = 3$$
but I am stuck at that point. Any hints?
 A: Just use definition. $\log_b A=c$ means $b^c=A$. Thus your question becomes.
$$x^2-5x-28=2^3$$
Now solve for $x$.
A: The first thing is that you can't distribute the $\log$, since it isn't a number, it's a function that doesn't obey linearity, i.e. $\log(\alpha+\beta)\neq\log\alpha+\log\beta$.
You can solve this problem this way: (move your mouse above each step to know which law was used)
$$\eqalign{\require{action}\log_2(x^2-5x-28)=3 \ \ &\Rightarrow \ \ \mathtip{2^{\log_2(x^2-5x-28)}=2^3}{\displaystyle\rm because\, a=b \iff c^a=c^b. } \\ &\Rightarrow \ \  \mathtip{x^2-5x-28=8}{\rm \displaystyle\rm because\, b^{\displaystyle\log_bc}=c. } \\ &\Rightarrow \ \  \mathtip{x^2-5x-36=0}{\text{adding $-8$ to both sides}}.}$$
The last equation can be solved using the quadratic formula. And after you get your solutions, check them.
A: Rather than distribute the log, exponentiate both sides:
$$2^{\left(\log_2(x^2 − 5x − 28)\right)} = 2^3$$
A: How exactly you distribute the log?
$\log A + \log B= \log AB$ but 
$\log(A+B) \ne \log A+\log B$
What you are looking at is a quadratic equation.
$x^2-5x-28=2^3$
Because, if $y=\log_2x$, then $2^y=x$
$x^2-5x-28-8=0$
$x^2-5x-36=0$
$(x-9)(x+4)=0$
A: $$
\log_2\text{something} = 3 \text{ if and only if }2^3=\text{something}.
$$
That will get you the solution.
