Respuesta :
First of all, we *do* have to put explanations, so it's good that you want them, keep this attitude going!
As for the excercise, we want to use the property that allows us to swap sum/subtractions of logarithm into multiplication/division inside them:
[tex] \log(a)+\log(b) = \log(ab),\quad \log(a)-\log(b)=\log\left(\dfrac{a}{b}\right) [/tex]
So, we want to move all logarithmic terms to the left hand side: add [tex]\log_2(2-x) [/tex] to both sides to get
[tex] \log_2(-x)+\log_2(2-x) = 3 [/tex]
Now we can use the property we just invoked:
[tex] \log_2(-x(2-x)) = 3 [/tex]
This is a good position, because every equation of the form
[tex] \log_a(b)=c [/tex]
can be solved by considering both sides as the exponents of the base of the logarithm:
[tex] \log_a(b)=c \iff a^{\log_a(b)}=a^c \iff b = a^c[/tex]
(I used the fact that exponential and logarithm are the inverse one of the other: [tex] a^{\log_a(b)}=b [/tex])
So, we can conclude
[tex] -x(2-x) = 2^3 \iff -2x+x^2 = 8 \iff x^2-2x-8=0 [/tex]
This is a standard quadratic equation, whose solutions are [tex] x=-2,\ x=4 [/tex] (I assume you are confident in solving this kind of equation, let me know otherwise)
Finally, we have to check if these solutions can actually be fed in the equation: we have
[tex] x=-2 \implies \log_2(2)=3-\log_2(2+2) \iff 1=3-2 [/tex]
which is true. On the other hand,
[tex] x=4 \implies \log_2(-4)=3-\log_2(2-4) [/tex]
But this expression can't be computed, because you can't compute the logarithm of a negative number. So, the only feasible solution is [tex] x=-2 [/tex]
