<s>[INST] Exponential and Logarithmic Functions - $\log_2(x^2-4x-28)= 2$ I got $x= \frac{45}{x-4} $
I know its wrong. Can someone show me how to solve.
Thanks
 [/INST] Problem: We need to solve for $x$ in the given equation
$$\log_2(x^2-4x-28)=2$$
Remember that if $\log_a(b)=y$, then $a^y=b$. In our equation, $a=2$, $b=x^2-4x-28$, and $y=2$. Therefore $x^2-4x-28=2^2=4$. This is easy to solve for.
$$x^2-4x-28=4$$
$$x^2-4x-32=0$$
Factor it
$$(x-8)(x+4)=0$$
$$\color{green}{\boxed{x=8, \ -4}}$$
Hope I helped!
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