The question given is:
[tex]\begin{gathered} 3-\log _2(2-x)=\log _2(-x-5) \\ \end{gathered}[/tex]
In order to solve the question, we need to be conversant with laws of logarithm.
[tex]\begin{gathered} \log _22=1^{}(\log _aa=1) \\ \text{Multiply both sides by 3} \\ 3\times\log _22=3\times1 \\ 3\log _22=3 \\ \log _22^3=3 \end{gathered}[/tex]
Let us now plug in this new expression for 3 into the question.
[tex]\begin{gathered} \log _22^3-\log _2(2-x)=\log _2(-x-5) \\ \text{but,} \\ \log _ab-\log _aC=\log _a(\frac{b}{c}) \\ \therefore\log _2\mleft(\frac{2^3}{2-x}\mright)=\log _2(-x-5) \\ \\ \text{Since,} \\ \log _ax=\log _ax\text{ (because x = x)} \\ \\ \mleft(\frac{2^3}{2-x}\mright)=-x-5 \end{gathered}[/tex]
Now, it is only a matter of solving for x. This is done below.
[tex]\begin{gathered} \frac{8}{2-x}=-x-5 \\ \text{ Multiply both sides by (2-x)} \\ \frac{8}{(2-x)}\times(2-x)=(-x-5)\times(2-x) \\ (-x-5)=-(x+5) \\ 8=-(x+5)(2-x) \\ 8=(x-2)(x+5)\text{ (Expand the bracket)} \\ 8=x^2+5x-2x-10 \\ \\ \therefore x^2+3x-10-8=x^2+3x-18=0 \end{gathered}[/tex]
Solving the quadratic equation:
[tex]\begin{gathered} x^2+3x-18=0 \\ x^2+6x-3x-18=0 \\ \text{Factorize} \\ x(x+6)-3(x+6)=0 \\ (x-3)(x+6)=0 \\ x-3=0\text{ or} \\ x+6=0 \\ \\ \therefore x=3,x=-6 \end{gathered}[/tex]
Now that we have the possible values of x, we also need to check if any of these solutions is not extraneous. We do this by substituting the values of x into the question.
[tex]\begin{gathered} \text{When x = 3} \\ 3-\log _2(2-3)=\log _2(-3-5) \\ 3-\log _2(-1)=\log _2(-8) \end{gathered}[/tex]
We cannot proceed further than this because the log of a negative number is not defined.
The plot of logarithm to base two is shown below to emphasize this point.
Therefore, x = 3 is not an answer since it is extraneous
Now, let us check for x = -6
[tex]\begin{gathered} 3-\log _2(2-(-6))=\log _2(-(-6)-5) \\ 3-\log _2(8)=\log _2(6-5) \\ 3-\log _22^3=\log _21 \\ 3-3\log _22=\log _21 \\ \log _22=1 \\ \text{From the graph above,} \\ \log _21=0 \\ 3-3(1)=0 \\ 0=0 \end{gathered}[/tex]
Therefore, x = -6 is a proper solution.
Thus, the final answer is:
x = -6
